Problem #1.
 See Exam Handout.
Solution.
 By the definition of the weak topology $\tau'$ on $H$, each $\Lambda\in H^{\star}$ is continuous, and $\tau'$ is the weakest topology which insures this by declaring $V(\Lambda,U):=\Lambda^{1}(U)$ to be a subbasic open set for every open $U\in\mathbb{R}$. Since we are working in a topological vector space, we can significantly refine this in order to make this weak topology more tractable. First note the easy fact that any mapping $f$ from a topological space $X$ into a metric space $Y_{d}$ is continuous if and only if there exists an open set $U_{\epsilon}$ corresponding to every $\epsilon>0$ such that $d_{Y}(f(x_{1}f(x_{2}))<\epsilon$ whenever $x_{1},x_{2}\in U_{\epsilon}$ (this is a standard fact proved in Math 121), and the even easier fact that $\Lambda0=0$ by linearity. Now $\mathbb{R}$ is a metric space whose origin has a local base given by the balls $B_{\epsilon}(0)$ for all $\epsilon>0$; consequently, the origin $0$ of $H$ has a weak local base $\mathscr{B}$ given by the collection of all finite intersections of the subbasic open sets
$$V(\Lambda,\epsilon):=\left\{x\in H:\Lambda x<\epsilon\;\text{for all}\; \Lambda\in H^{\star}\;\text{and}\;\epsilon>0\right\},$$
and since translation on any topological vector space is continuous (Definition 1.9.1), this local base completely determines the weak topology $\tau'$ on all of $H$ (thus we can reduce most arguments which are strictly topological in nature to arguments concerning only the origin). Explicitly, every weak neighborhood $U$ of $0$ is the union of basic open sets of the form
$$V=\bigcap\limits_{j=1}^{n}V(\Lambda_{j},\epsilon_{j})=\{x\in H:\Lambda_{j} x<\epsilon_{j}\;\text{for every}\;1\leq j\leq n<\infty\},$$
and after a translation by $x$, each weak neighborhood of a point $x\in H$ is the union of the basic open sets of the form $x+V$. (The notation $\Lambda_{j}$ is for convenience, and in no way implies we are ordering or otherwise enumerating $H^{\star}$.) Fixing $U$ and taking the intersection of the kernels of each $\Lambda_{j}$ (which is a subspace of $H$) corresponding to some basic open set $V$ which $U$ contains, we find
$$\{0\}\subset K:=\bigcap\limits_{j=1}^{n}\text{Ker}(\Lambda_{j})=\{x\in H:\Lambda_{j}=0<\epsilon;\text{for all}\;1\leq j\leq n\}\subset V\subset U,$$
and since $x\mapsto (\Lambda_{1},\ldots,\Lambda_{n})$ takes $X$ into $\mathbb{R}^{n}$ with kernel $K$ as above, it follows that
$$\dim X\leq \dim\mathbb{R}^{n}+\dim K,$$
this implying (since $H$ is infinite dimensional) $\dim K=\infty.$ Consequently, every weak neighborhood $U$ of $0$ contains an infinite dimensional subspace of $H$! Among other things, this proves that $U\cap S\neq\emptyset$ since clearly $U$ contains every point of the form $\alpha y$, where $\alpha\in\mathbb{R}$ and $y\in K$, and in particular the point $\frac{y}{y}\in S.$ Translating the origin to any point $x\in H$ (cf. remark above) then shows that $U_{x}$ is a weak neighborhood of $x$ and that $U_{x}\supset K_{x}$, i.e. every weak neighborhood $U_{x}$ of every point $x\in H$ contains the infinite dimensional affine subspace $K_{x}$, and thus a line segment of the form $x+\alpha y$ passing through $x$. It follows that $U_{x}\cap S\neq\emptyset$ whenever $x\leq1$, since in this case we can find an $\alpha$ corresponding to any $y\in K_{x}$ such that $x+\alpha y=1$ (for Hilbert spaces we may, in applying the Pythagorean identity, take $\alpha=\left(\frac{1x^{2}}{y^{2}}\right)^{\frac{1}{2}}$ by choosing $y\in K_{x}$ orthogonal to $x$). We conclude therefore that $\overline{S}^{w}\supset B$ since every weak neighborhood of every $x\in B$ contains an infinite dimensional subspace which intersects $S$.
(See remark (1) below) To obtain the reverse inclusion, we must exhibit, for each $x\in H$ with $x>1$, a weak neighborhood $U$ which does not intersect $S$. In the spirit of the preceding geometric proof, the geometric HahnBanach theorem (Theorem 1.5.14) easily provides us with such a neighborhood (see remark (4) below for an alternative approach to applying it). To apply it, we first record the obvious facts that the strong (norm) topology on $H$ is locally convex (Exercises 1.3.3 and 1.9.1), that it is normal (T4) in the sense that its basic open sets separate closed sets (this easy fact about metric spaces is standard in basic topology), and that $B$ is strongly closed; hence, there is a convex strongly open neighborhood $W$ of $\{x\}$ such that $W\cap B=\emptyset$. In light of the convexity of $B$, all of the conditions of the geometric Hahn Banach theorem are satisfied, and so there is a hyperplane in $H$ which separates $B$ and $W$. Precisely, for each $x\notin B$, there exists a linear functional $\Lambda\in H^{\star}$ and $c\in\mathbb{R}$ such that $$\Lambda x_{0}< c\leq\Lambda z$$ for every $x_{0}\in W$ and $z\in B.$ By definition the pullback $\Lambda^{1}(\infty,c)$ is a weak neighborhood $U$ of $x$, and by the separation bound above $U\cap B=\emptyset.$ $U$ is then a weak neighborhood of $x$ which does not intersect $S$, and so $x\notin\overline{S}^{w}$, i.e. $\overline{S}^{w}\subset B^{c}$. The claim follows by combining the two inclusions.
Remarks:
(1) After concluding $\overline{S}^{w}\supset B$, we can use Exercise 1.9.20 to conclude $\overline{S}^{w}=B$, since this exercise shows $B$ is weakly closed. This seemed to violate the instructions of the exam though, since the exercise is too closely related; in any case, the proof above provided a nice application of some of the optional material in the text (namely, the geometric HahnBanach theorem), and facilitates a slight generalization to the problem as indicated below.
(2) Note that the above proof goes through when $H$ is merely an infinitedimensional normed space (not necessarily Banach); moreover, since every continuous linear functional $\Lambda$ is completely determined by its continuous real part (recall $z=\Re zi\Re(iz)$ for every $z\in\mathbb{C}$), the geometric HahnBanach theorem continues to hold in a complex topological space $X$, with the separation being given by $\Re\Lambda$ and $c\in\mathbb{R}$. Indeed, with the real case already proved, it is easy to see that the so constructed continuous real linear functional $\lambda$ is the real part of a unique continuous complex linear functional $\Lambda$, and that $\Lambda x=\lambda xi\lambda(ix)$. Thus, the conclusion can be generalized to: the weak close of the unit sphere in an infinite dimensional complex normed space is the unit ball.
(3) Note that an immediate implication of the result above is that the strong topology strictly contains the weak topology in an infinite dimensional normed space, since $S$ is closed in the norm topology. Here's another observation that implies the same result, among others. In the proof above we showed that every weak neighborhood of $0$ contains an infinite dimensional subspace of $H$. This implies that the weak topology on $H$ is never locally bounded whenever $H$ is infinitedimensional, regardless of whether $H$ is normed, metrized, etc.; thus, if $H$ is a locally bounded and infinite dimensional topological vector space (note that normed spaces are locally bounded), then the strong topology on $H$ strictly contains the weak topology on $H$. Taken together, these observations provide a nice extension to part of Exercise 1.19.13. Thus, insofar as $H$ is normed, the weak and strong topologies never coincide when $H$ is infinite dimensional. On the other hand, we proved (after making some minor adjustments) that at least the weak and strong closures of any convex set coincide. There are numerous other observations in this direction; for example, the weak topology on $H$ is never locally compact (despite the unit ball being weakly compact when $H$ is reflexive) whenever $H$ is infinite dimensional (Exercise 1.9.24); it is also straightforward to prove the converse of Exercise 1.9.24, thus showing a topological vector space is locally compact if and only if it is finite dimensional, thereby implying that the weak and strong topologies coincide whenever $H$ is finitedimensional, since both topologies are Hausdorff, and one cannot weaken or strengthen such a topology without losing the Hausdorff or local compactness property, respectively.
(4) There is a very convenient form of the geometric HahnBanach theorem which is stated on its Wikipedia page, in that if the topological vector space $X$ is locally convex, $A$ compact, and $B$ closed, then there exists a hyperplane separating these sets with an even stronger condition than in the case for when $A$ is open and both $A$ and $B$ are convex. Specifically, there are two scalars $c,d\in\mathbb{R}$ and a continuous real linear functional $\Lambda\in H^{\star}$ (cf remark (2) for the complex case) such that for all $x\in A$ and $y\in B$, $$\Lambda x< c< d<\Lambda y$$ holds. Then in the proof above, one can take $\{x\}\in B^{c}$ to be the compact set $A$ and the ball $B$ to be the closed set $B$. Since this form of the HahnBanach theorem is not needed, I will not prove it in detail, but its proof is fairly obvious and consists of recognizing that if $V$ is a convex neighborhood of $0$ (which exists since $X$ is now assumed locally convex), then $A+V$ is open and convex (the convexity is obvious; that it is open when $A$ is compact follows from $A+V=\cup_{x\in A}x+V$, i.e. a union of open sets index by elements in $A$), and one can arrange so that $(A+V)\cap B=\emptyset$ (this is a sort of ``almost'' normal property that all topological vector spaces have, in that a pair of compact and closed sets can be separated by open sets, but not necessarily two closed sets). Then one simply applies the original geometric HahnBanach theorem to the sets $A+V$ and $B$ to obtain a separating hyperplane $\Lambda$, noting then that $\Lambda(A)\subset\Lambda(A+V)$ and that $\Lambda(A)$ is compact in $\mathbb{R}$ and separated from $\Lambda(B)$.
(5) While this remark is not directly related to the problem above, it seems worthwhile to mention it in light of remark (4). From the modified geometric HahnBanach theorem, we see that in a locally convex space the dual space $X^{\star}$ always separates points on $X$. But this is the only essential ingredient required to prove that the weak topology on $X$ is Hausdorff, thus providing a generalization to Exercise 1.9.14 (that the weak* topology on $X^{\star}$ is Hausdorff follows immediately from the obvious fact that the linear functionals $\lambda_{x}(\Lambda)=\Lambda(x)$ are continuous (by definition) and separate points in $X^{\star}$).
Problem #2.
 See Exam Handout.
Solution.

Let $\tau$ and $\upsilon$ denote the strong (norm) topologies on $X$ and $Y$, respectively, with a $'$ indicating the corresponding weak topologies. Then if $T$ is continuous from $X_{\tau'}\to Y_{\upsilon'}$, it is obviously continuous from $X_{\tau}\to Y_{\upsilon'}$ since $\tau\supset\tau'$ (Exercise 1.9.13). But $Y_{\upsilon'}$ is Hausdorff (Exercise 1.9.14), $\upsilon\supset\upsilon'$, and so $T$ is continuous from $X_{\tau}\to Y_{\upsilon}$ as a consequence of the closed graph theorem (Theorem 1.7.19, condition (iii)). Suppose now $T$ is continuous from $X_{\tau}\to Y_{\upsilon}$ and let $U$ be a weakly open set in $Y_{\upsilon'}$. By the definition of the weak topology, there is a corresponding collection of open sets $V_{i}\subset\mathbb{C}$ such that
$$W:=T^{1}(U)=T^{1}\left(\cup_{\alpha}\cap_{i}\Lambda_{i}^{1}(V_{i})\right)=\cup_{\alpha}\cap_{i}T^{1}\left(\Lambda_{i}^{1}(V_{i})\right)=\cup_{\alpha}\cap_{i}\left(\Lambda_{i}\circ T\right)^{1}(V_{i}),$$
where $i=1,2,\ldots,n$, $\alpha\in A$ is some (arbitrary) index, and $\Lambda_{i}\in Y^{\star}$. The continuity of $T$ from $X_{\tau'}\to Y_{\upsilon'}$ now follows from showing $W$ is weakly open in $X_{\tau'}$, and this will follow by showing the continuity of each $\lambda_{i}:=\Lambda_{i}\circ T$ when viewed as maps from $X_{\tau'}\to Y_{\upsilon'}\to\mathbb{C}$. To that end, note first the obvious fact that $T$ is continuous from $X_{\tau}\to Y_{\upsilon'}$ since $\upsilon\supset\upsilon'$. Then since the composition of two linear continuous functions is again a linear continuous function (they form a type of algebra), we see $\lambda$ as defined above for arbitrary $\Lambda\in Y^{\star}$ is a linear continuous map from $X_{\tau}\to Y_{\upsilon'}\to\mathbb{C}$, i.e. $\lambda\in X^{\star}.$ But then by definition of the weak topology again, $\lambda$ is also a linear continuous map from $X_{\tau'}\to Y_{\upsilon'}\to\mathbb{C}$, and the claim follows from the preceding observations.
Problem #3.
 See Exam Handout.
Solution.

By assumption, for every $x\in H$ we have
$$0\leqT^{n+1}x=TT^{n}x\leqT_{\text{op}}T^{n}x\leqT^{n}x,$$
and
$$T^{n}x\leqT^{n}_{\text{op}}x=T_{\text{op}}^{n}x\leqx<\infty.$$
(The relation $T^{n}=T^{n}$ is a consequence of $T$ being selfadjoint since $T^{*}T=T^{2}$ for any $T\in B(H)$). Thus, being a bounded monotonically decreasing sequence, $\lim_{n\to\infty}T^{n}x$ exists for every $x\in H$. We now compute the Cauchy difference of the sequence $T^{2n}x$ to obtain
\begin{align*}
T^{2n}xT^{2m}x^{2}
&=(T^{2n}xT^{2m}x\;\;T^{2n}xT^{2m}x)\\
&=(T^{4n}x\;\;x)2(T^{2(n+m)}x\;\;x)+(T^{4m}x\;\;x)\\
&=T^{2n}x^{2}2(T^{2(n+m)}x\;\;x)+T^{2m}x^{2}.
\end{align*}
Since $$T^{n}x^{2}=(T^{n}x\;\;T^{n}x)=(T^{2n}x\;\;x)$$ converges as $n\to\infty$, we find that the subsequences $(T^{4n}\;\;x)$, $(T^{4m}x\;\;x)$, and $(T^{2(m+n)}\;\;x)$ all converge to the same limit (note $2(m+n)$ is even, so is a subsequence of a convergent subsequence, cf. Exercise 1.6.8). Thus, $T^{2n}xT^{2m}x\to0$ as $m,n\to\infty$. This holds for every $x$, so we conclude $\lim_{n\to\infty}T^{2n}$ exists in the strong operator topology in $B(H)$.

Since $x\in V$ if and only if $T^{2}xx=0$, we see that $V=\text{Ker}(T^{2}I)=(T^{2}I)^{1}(\{0\})$, which is a closed subspace of $H$ since $T^{2}I$ is continuous and the singleton set $\{0\}$ is closed. Now let $Tx=\lim_{n\to\infty}T^{2n}x$, which exists by (a) for every $x\in H$. It is clear that $T$ is linear because limit operations are linear and each $T^{2n}$ is. That $T$ is bounded is obvious from its definition, thus continuous $T$ is continuous and $T\in B(H)$. To show that $T$ is an orthogonal projection of some subspace of $H$, it is enough to show that $T$ is an idempotent selfadjoint operator. To that end, fix $x,y\in H$ and compute
$$(x,T^{*}Ty)=(Tx\;\;Ty)=\lim\limits_{n\to\infty}(T^{2n}x\;\;T^{2n}y)=\lim_{n\to\infty}(x\;\;T^{2n^{*}}T^{2n}y)=\lim_{n\to\infty}(x\;\;T^{4n}y)=(x\;\;Ty),$$
and also in the same manner
$$(Tx,Ty)=\lim_{n\to\infty}(T^{4n}x\;\;y)=(Tx\;\;y).$$
The conclusions that $T=T^{*}T=TT^{*}=T^{2}$ all follow immediately from this; thus, $T=\pi_{S}$ for some subspace $S=\mathscr{R}(T)$. Suppose now $x\in V$. Then $T^{2}x=x$, so $T^{2n}x=x$ for each $n$, and thus $Tx=x$, the identity operator on $V$. We have that $S\supset V$, and to show the reverse inclusion, take $x\in H$ and let $\pi_{V}$ be the orthogonal projection of $V$. We need to show $\pi_{V}Tx=Tx$ so that we have $Tx\in V$, hence $S\subset V$.

I wasn't able to come up with a counterexample in time; mostly because I don't have any intuition on what a selfadjoint operator is, let alone many concrete examples (barring projections and the integration operator in problem 6). But here's a quick example that everyone is familiar with from Fourier series that nicely illustrates how hard it is to converge in the uniform topology. Consider the space $L^{2}([0,2\pi])$ and the orthogonal projection operators $\pi_{i}f=(f\;\;e^{int})e^{int}$ onto the orthonormal basis $\{e^{int}\}_{i=1}^{\infty}.$. Obviously each $\pi_{i}$ is selfadjoint (being projections), and it is clear each $\pi_{i}=1$ by their definitions for each $i$. The Parasavel identity shows that $\pi_{i}\to0$ in the strong topology, since $f^{2}=\sum_{i=1}^{\infty}\pi_{i}f^{2}$ so that we must have $\pi_{i}f\to0$ as $i\to\infty$ for each $f\in L^{2}$. But $\pi_{i}=1$, so by uniqueness of limits, we conclude $\pi_{j}$ does not converge in the uniform topology.
Problem #4.
 See Exam Handout.
Solution.
 (Note: This ended up being one of the more difficult problems on the exam, and I have kept my original line of thought mostly in tact to illustrate this (mostly for myself as a learning aid)).
Since we need to show strong convergence, let's begin by fixing $x\in H$ and computing the Cauchy difference (with $T_{n}:=(\pi_{V}\pi_{W})^{n}$) \begin{align*} T_{n}xT_{m}x^{2} &=\left(T_{n}xT_{m}x\;\;T_{n}xT_{m}x\right)\\ &=\left(T_{n}x\;\;T_{n}x\right)2\left(T_{n}x\;\;T_{m}x\right)+\left(T_{m}x\;\;T_{m}x\right)\\ &=\left(T_{n}^{*}T_{n}x\;\;x\right)2\left(T_{m}^{*}T_{n}x\;\;x\right)+\left(T_{m}^{*}T_{m}x\;\;x\right)\\ &=\left(\pi_{W}T_{2n1}x\;\;x\right)2\left(\pi_{W}T_{n+m1}x\;\;x\right)+\left(\pi_{W}T_{2m1}x\;\;x\right)\\ &=T_{n}x^{2}2\left(T_{n}x\;\;T_{m}x\right)+T_{m}x^{2}. \end{align*} The second to last line follows from the identity $$T_{n}^{*}=(\pi_{V}\pi_{W}\ldots\pi_{V}\pi_{W})^{*}=\pi_{W}^{*}\pi_{V}^{*}\ldots\pi_{W}^{*}\pi_{V}^{*}=\pi_{W}\pi_{V}\pi_{W}\ldots\pi_{V}\pi_{W}\pi_{V}=\pi_{W}T_{n1}\pi_{V}$$ (where the selfadjointness of each projection is used) and then applying the idempotent identity $\pi_{V}^{2}=\pi_{V}$ to rewrite the resulting composition of operators; the last line, which is true for any squared inner product, is written for comparison to the more interesting line before. Since $\pi_{W}$ is an orthogonal projection, it follows that $$T_{n}x^{2}=(\pi_{W}T_{2n1}x\;\;x)\leq(T_{2n1}x\;\;x),$$ this being because if $y\in H$ and $\pi_{S}$ is an orthogonal projection onto a closed subspace $S$, then $y=\pi_{S}y+\pi_{S^{\perp}}y$ and so by the Pythagorean identity $y^{2}=\pi_{S}y^{2}+\pi_{S^{\perp}}^{2}$, i.e. $\pi_{S}y\leqy$ (one then just uses the polarization identity to conclude the same thing for inner products). Consequently, $$T_{n+1}x^{2}\leq(T_{2n+1}x\;\;x)=(\pi_{V}\pi_{W}\pi_{V}\pi_{W}T_{2n1}x\;\;x)\leq(\pi_{W}T_{2n1}x\;\;x)=T_{n}x^{2}\;(!)$$ Thus, being a monotonically decreasing sequence of nonnegative numbers, $\lim\limits_{n\to\infty}T_{n}x$ exists for every $x\in H$ since $H$ is complete. But this does not show $\lim_{n\to\infty}T_{n}x$ exists! We need some kind of weak convergence before we can make this claim (cf. Exercise 1.9.18 for instance). But at least notice that our remarks above also show that the subsequence $\{(T_{j}x\;\;x)\}_{j\;\text{odd}}$ converges, since this is also a monotonically decreasing sequence bounded below by $0$. Suppose now we could arrange $n+m1$ to always be odd, that is, of the form $2j1$ (where $j\to\infty$ as $m,n\to\infty$). Then we could have $$T_{n}xT_{m}x^{2}\leq(T_{2n1}x\;\;x)+(T_{2m1}x\;\;x)2(T_{2j1}x\;\;x)\leq(T_{2n+1}x\;\;x)+(T_{2m+1}x\;\;x)2(T_{2j+1}x\;\;x)\leq\ldots.$$ We know that the right hand side of this inequality tends to $0$ as $m,n\to\infty$; however, because of the inequality, we cannot directly conclude $T_{n}xT_{m}x\to0$ as $m,n\to \infty$. We could fix this by carrying around a chain of extra projections as indicated above, but this is awkward, and obviously does not resolve the more immediate issue of needing $m+n1$ to be odd. But suppose we could show $x'=\lim_{n\to\infty}Tx$ existed for every $x\in H$. Then we would immediately see that $T$ was linear because limit operations are linear and each $T_{n}$ is; moreover, if $x\in V\cap W$, then $T_{n}=(\pi_{V}\pi_{W}\ldots\pi_{V}\pi_{W})x=x$, so $Tx=x$, and $T$ would be the identity on $V\cap W$ as required. But even if we could get here, we have only concluded upto this point that $T=\pi_{V\cap W}$ for $x\in V\cap W$. To finish the proof, we would need to show that $T$ is an orthogonal projection (e.g. selfadjoint, idempotent, etc.) and that $\mathscr{R}(T)\subset W\cap V$ (the reverse inclusion $\mathscr{R}(T)\supset W\cap V$ was already shown). But working with the sequence $T_{n}$ leads to further difficulties; for example, in an attempt to show that $T$ is selfadjoint, we would write (using the continuity of the inner product) $(Tx,y)=\lim_{n\to\infty}(T_{n}x\;\;y)=\lim_{n\to\infty}(x\;\;\pi_{W}T_{n1}\pi_{V}y)=(x\;\;\pi_{W}T\pi_{V}y),$ and thus would need to verify that $\pi_{V}T\pi_{W}y=Ty$. To show the inclusion $\mathscr{R}(T)\subset V\cap W$, one obvious way to proceed would be by verifying $\pi_{W}Tx=Tx=\pi_{V}Tx$, so that for each $x\in H$ we would have $Tx\in V\cap W$. Some of these equalities are trivial; for example, since $\pi_{V}T_{n}x=T_{n}x$ for all $x,n$ and $\pi_{V}$ is obviously bounded and continuous, we could conclude by taking limits that $\pi_{V}Tx=Tx\in V$ (cf. Exercise 1.9.29). The same method shows $T\pi_{W}x=Tx$, but this equality does not help us, and the method fails for $\pi_{W}T_{n}x$ since we cannot apply any idempotent relation here. We conclude that this direct approach is fruitless.
In light of the above observations, there is a clear, though admittedly awkward, way of resolving our issues (although I would be interested in seeing if the direct approach above could be resolved). Our major source of frustration stemmed from the need to ensure $m+n1$ was odd in the Cauchy difference above, and also from the lack of control in applying an indempotent relation. We can resolve this by considering the recursively defined sequence $T'_{1}:=\pi_{V}$, $T_{2}=\pi_{W}\pi_{V}$, $T'_{3}=\pi_{V}\pi_{W}\pi_{V}$, $\pi_{W}\pi_{V}\pi_{W}\pi_{V}$, $\ldots$, which is similar to the sequence $T_{n}$. A quick glance at our previous work then shows the above relations continue to hold and with a couple of notational changes, we can prove the above assertions which we had difficulty with for $T_{n}$. However, even once this is done, it is still possible that $T'_{n}x\neq T_{n}x$ as $n\to\infty$ for each $x\in H$ ($T'_{n}$ is not even a subsequence of $T_{n}$, and we still have not shown $T_{n}$ even converges). However, this is easily corrected by showing that the ``reverse'' sequence $T''_{1}=\pi_{W}$, $T''_{1}=\pi_{V}\pi_{W}$, $T''_{3}=\pi_{W}\pi_{V}\pi_{W}$, $\ldots$ (for which $n$ even we have $T'_{n}=T''^{*}_{n}$ and for $n$ odd each term in each sequence is selfadjoint) has the same strong limit as $T'$. Once this is done, we will then be able to conclude $\lim_{n\to\infty}T_{n}$ exists and that $\lim T_{n}=\lim T_{n}'=\lim_{n}T_{n}''=T.$
We now proceed to show $T_{n}'$ and $T_{n}''$ both converge strongly to a common limit $T=\pi_{V\cap W}$; this will then finally show $T_{n}$ converges strongly to $\pi_{V\cap W}$, and the proof will be complete. In that vein, we begin with $T'$ and proceed exactly as before by fixing $x\in H$ and examining the convergence of the sequence $T'_{n}x$. We first note the similar (though different) adjoint relations $$T'^{*}_{n}T'_{n}=T'_{2n1},$$ and $$T'^{*}_{n}T'_{m}=T'_{m+nl},$$ where the integer $l$ is needed because of the differences which can occur when $m$ and $n$ have the same or opposite parity (even/oddness). When they are both odd or even, $l=1$, and when one is odd and the other even, $l=0$. Note that we no longer have extra projections to worry about, and so we can maintain equality in the Cauchy difference below! Note also that now $m+nl$ is always odd (addition assigns even parity to terms which are the same, and assigns odd parity to terms which differ), and changes it when they are different), so can be written as $2j11$ for some $j(m,n)\to\infty$ as $m,n\to\infty$. Thus, we find exactly as before (and making the above changes) that $$T'_{n}xT'_{m}x^{2}=(T'_{2n1}x\;\;x)2(T'_{2j1}x\;\;x)+(T'_{2m1}x\;\;x).$$ In the same manner, we deduce that $$(T'_{2n1}x\;\;x)=T'_{n}^{2}$$ so $$(T'_{2n+1}x\;\;x)=T'_{n+1}^{2}=\pi_{V}T'_{n}^{2}\leqT'_{n}^{2}=(T'_{2n1}x\;\;x)$$ or $$(T'_{2n+1}x\;\;x)=T'_{n+1}^{2}=\pi_{W}T'_{n}^{2}\leqT'_{n}^{2}=(T'_{2n1}x\;\;x)$$ depending on whether $n$ is odd or even. In either case, we find that $\lim_{n\to\infty}(T'_{2n1}x\;\;x)$ exists, since this subsequence (indexed by odd numbers) is bounded below by $0$ and nonincreasing. Therefore, we conclude at last $$\lim\limits_{n\to\infty}T'_{x}=x'\in H$$ for every $x\in H$, since $$T'_{n}xT'_{m}x\to0$$ as $m,n\to\infty$. Now we finish the proof. $T'$ is linear since limit operations are, and each $T'_{n}$ is. It is also clear that $\mathscr{R}(T')\subset V\cap W$ and that on $V\cap W$ $T'$ is the identity since for such $x$ we have $\pi_{V}x=x$, $\pi_{W}x=x$, so that $T'_{n}x=x$ for each $n$. One of the motivating reasons for the definition of the sequence $T'_{n}$ is so that we could prove $\pi_{V}T'=T'$ and $\pi_{W}T'=T'$. Indeed, $\pi_{V}T'_{2n}x=T'_{2n+1}x$ and $\pi_{W}T'_{2n1}x=T'_{2n}x$ by the relations above. Therefore, we can take $n\to\infty$ and conclude $\pi_{V}T'x=\pi_{W}T'x=T'x$ for all $x\in H$ (cf. Exercise 1.9.29; note that we have also used the fact that if a sequence converges, all of its convergent subsequences converge to the same limit). It follows that $\mathscr{R}(T')\subset V\cap W$ since $T'x=\pi_{V}T'x\in V$ and $T'x=\pi_{W}T'x\in W$, so $T'x\in V\cap W$ for every $x\in H$. Consequently, $\mathscr{R}(T')=W\cap V$, and $T$ is the identity on $W\cap V$. It remains to show that $T'$ is an orhtogonal projection, and this will follow from showing $T'$ is indempotent and selfadjoint. If $x,y\in H$, then the continuity of the inner product and our computations above show that $$(x\;\;T'^{*}T'y)=(T'x\;\;T'y)=\lim_{n\to\infty}(T'_{n}x,\;\;T'_{n}y)=\lim_{n\to\infty}(T'_{2n1}x\;\;y)=(T'x\;\;y)=(x\;\;T'^{*}y),$$ and also (in the same manner) $$(T'x,T'y)=\lim_{n\to\infty}(x\;\;T'_{2n1}y)=(x\;\;T'y).$$ All of the relations $T'=T'^{*}T'=T'T'^{*}=T'^{2}$ follow immediately from these these results; thus, $T'=\pi_{V\cap W}$. Since it is obvious that the exact same argument goes through with $T''_{n}$, we conclude $T'=T''$ and so $$\lim_{n\to\infty}Tx=\lim_{n\to\infty}T'x=\lim_{n\to\infty}T''x=\lim_{n\to\infty}(\pi_{V}\pi_{W})^{n}x=\pi_{V\cap W}x$$ for every $x\in H$; that is, $(\pi_{V}\pi_{W})^{n}\to\pi_{V\cap W}$ strongly in $B(H\to H)$. }
Problem #5.
 See Exam Handout.
Solution.
 Following the hint, a quick mental calculation motivates us to define linear functionals
$$\Lambda_{n}^{x}(f):=(f*K_{n})(x)=\int_{\mathbb{R}}K_{n}(xy)f(y)\;dx,$$
since then
$$\Lambda_{n}^{x}_{op}=\sup\limits_{f\in BC(\mathbb{R}\to\mathbb{R}),\;f_{L^{\infty}}=1}\left\int_{\mathbb{R}}K_{n}(xy)f(y)\;dy\right\leq\int_{\mathbb{R}}K_{n}(y)\;dy=K_{n}_{L^{1}}.$$
But this bound is only useful if we can demonstrate equality, or otherwise modify it to obtain a lower bound, since we are attempting to upper bound the term $K_{n}_{L^{1}}.$ To that end, fix $x=0$ and consider just the set of linear functionals $\{\Lambda_{n}\}.$ For fixed $n$, it is clear that the supremum defining $\Lambda_{n}$ above will be achieved by the continuous function which is most frequently $1$ in absolute value and of the same sign as $K_{n}$. Absent continuity, such a function is the simple function $\psi_{n}$ defined by $\psi_{n}(y)=1$ when $K_{n}(y)\geq0$ and $\psi_{n}(y)=1$ whenever $K_{n}(y)<0
Remarks. This exercise complements Example 1.7.9 following the proof of the uniform boundedness principle, concerning the Fourier transform. In fact, in Math 136 we showed that the Dirichlet kernels $\{D_{n}\}_{n=1}^{\infty}$ for the partial sums of the Fourier series of continuous functions on $[0,2\pi]$ have divergent $L^{1}$ norm as $n\to\infty$. If we replace $K_{n}$ by $D_{n}$ in this problem, then we are in the exact opposite situation since in this case we have an unbounded set of linear functionals, and the uniform boundedness principle would imply then that for every $x$ in $[0,2\pi]$ that there is a corresponding dense collection of continuous functions whose Fourier series diverge at $x$!
Problem #6.
 See Exam Handout.
Solution.

First note that
$$Tf(x)=\left\int_{0}^{x}f(y)\;dy\right\leq\int_{0}^{x}f(y)\;dy\leqf_{1}\leq\chi_{[0,1]}_{2}f_{2}<\infty$$
by Holder's inequality and the fact that the indicator function $\chi_{[0,1]}$ has norm $1$ on $L^{2}([0,1])$. This shows that $T$ is welldefined, which isn't immediately obvious since $f$ is only assumed to be in $L^{2}([0,1])$, not $L^{1}([0,1])$ (and in fact this is only true because $[0,1]$ is finite, a counterexample being $x^{1}$ on $[1,\infty]$  thus, we see another curious instance of the interplay between containments of $L^{p}$ spaces and how such containments depend not only on the exponents $p$ and $q$, but also on the underlying support of each function; see Exercises 1.3.1114). Let
$$\mathscr{F}:=\{Tf:f_{2}\leq1\},$$
and observe for each $F\in\mathscr{F}$, $x,y\in[0,1]$, and $x\geq y$ that
$$F(x)F(y)\leq\int_{y}^{x}f(t)\;dt=\int_{0}^{1}f(t)\chi_{[x,y]}(t)\;dt\leq\chi_{[x,y]}_{2}f_{2}\leq Axy^{\alpha}=xy^{\frac{1}{2}}.$$
Thus, $T$ takes the unit ball $B(0;1)\subset L^{2}([0,1])$ into a subset $\mathscr{F}$ of the Holder space $H^{1,\frac{1}{2}}([0,1])$. It is clear that each $F\in\mathscr{F}$ is continuous on the compact set $[0,1]$, and that the family $\mathscr{F}$ is equicontinuous with continuity mode $\delta=\epsilon^{2}$ (owing to the uniform Holder continuity for each $F$) and uniformly bounded by $1$. All of the conditions of the ArzelaAscoli theorem are satisfied (Theorem 1.8.23) and we conclude $\overline{\mathscr{F}}$ is compact in $BC([0,1]\to\mathbb{R})$; that is,$$\mathscr{F}=T(B(0;1))\subset H^{1,\frac{1}{2}}[0,1]\subset BC([0,1])\subset L^{2}([0,1])$$ is precompact in $BC([0,1])$. It remains to show that this precompactness in $BC([0,1])$ implies precompactness in the norm topology of $L^{2}([0,1])$. The easiest way to show this is by using the sequential version of the ArzelaAscoli theorem (Corollary 1.8.25). Consider any sequence $\{F_{n}\}$ in $\mathscr{F}$. Then such a subsequence has a convergent subsequence $F_{n_{j}}$ which converges uniformly on $[0,1]$ as $j\to\infty$ to a limit $F$, i.e. converges in the $L^{\infty}$ norm. But observe that
$$\lim_{j\to\infty}F_{n_{j}}F_{2}=\lim_{j\to\infty}\int_{0}^{1}F_{n_{j}}(x)F(x)^{2}\;dx=0$$
by the dominated convergence theorem (actually, dominated convergence is not needed here, since the underlying measure space $[0,1]$ is finite, every $f\in L^{\infty}$ is automatically in $L^{2}$; in particular, estimating the integral after taking $j$ so large that $f_{n_{j}}(x)F(x)<\epsilon$ for all $x\in[0,1]$ gives $f_{n_{j}}f_{2}<\epsilon^{2}$), and so $\mathscr{F}$ is precompact in $L^{2}$ since sequential compactness is equivalent to compactness in normed spaces. Note that the crucial fact here is that the measure space $[0,1]$ is finite, and Exercise 1.3.11 elaborates on this. In any case, we see that $T$ is compact as required.

My immediate conjecture was the differentiation operator! But alas, a simple computation shows $T^{*}$ is not nearly this exciting. Indeed, for $f,g\in L^{2}([0,1])$ we need $(Tf\;\;g)=(f\;\;T^{*}g),$ so following our nose we compute
\begin{align*}
(Tf\;\;g)
&=\int_{[0,1]}\left(\int_{[0,x]}f(y)\;dy\right)\overline{g(x)}\;dx\\
&=\int_{[0,1]}\overline{g(x)}\int_{[0,1]}f(y)\chi_{[0,x]}(y)\;dydx\\
&=\int_{[0,1]}f(y)\int_{[0,1]}\overline{g(x)}\chi_{[0,y]}(x)\;dxdy\\
&=\int_{[0,1]}f(y)\overline{\int_{[0,1]}\overline{\chi_{[0,y]}(x)}g(x)\;dx}dy.
&=\int_{[0,1]}f(y)\overline{\int_{[0,1]}\chi_{[0,y]}(x)g(x)\;dx}dy.
\end{align*}
Thus, taking $T^{\star}=T$ works, and by uniqueness of the adjoint, we find that this particular integral operator is selfadjoint. Note that the use of Fubini's theorem to change the order of integration is justified since it is clear $F(x,y):=\overline{g(x)}f(y)\chi[0,x](y)$ is absolutely integrable on $[0,1]\times[0,1]$ as $f_{1}\leq1_{2}f_{2}=f_{2}<\infty$ and $\bar{g}_{1}\leq1_{2}\bar{g}_{2}=\bar{g}_{2}<\infty$ by Holder's inequality.
Problem #7.
 See Exam Handout.
Solution.

Inspired by the similarities of semicontinuous functions with jump functions, define $\{D_{n}\}_{n=1}^{\infty}$ by
$$D_{n}:=\left\{x\in X:\omega_{f}(x)>\frac{1}{n}\right\},$$
where $\omega_{f}(x)=\lim_{r\to0}\Omega_{f}\left(B(x;r)\right)$ is the oscillation of $f$ at $x$ (cf Math 131). Note that $D:=\bigcup_{n}D_{n}\subset X$ is the set of discontinuities of $f$ in $X$ since $f$ is continuous at $x$ if and only if $\omega_{f}(x)=0$ (cf Math 131). Now $f$ is upper semicontinuous and $X$ is complete (hence normal by Exercise 1.10.1), so there exists continuous functions $f_{j}:X\to(\infty,\infty]$ such that $f_{j}(x)\geq f(x)$ and $f_{j}(x)\to f(x)$ as $j\to\infty$ for every $x\in X$ (Exercise 1.10.11). Now choose any $N>0$ and any ball $B$ in $X$ and define sets
$$E_{m}=\bigcap\limits_{i,j=m}^{\infty}\left\{x\in B:d(f_{i}(x)f_{j}(x))\leq\frac{1}{2N}\right\}.$$
Then $B=\bigcup_{m}E_{m}$ since $f_{j}\to f$ for every $x$; thus, by Baire's category theorem (Theorem 1.7.3), there is at least one $E_{M}$ which is dense in a subball $B'\subset B$. Moreover, the continuity of each $f_{j}$ implies each $E_{m}$ is the intersection of closed sets, thus $E_{M}$ is closed and we have $E_{M}=B'$. This implies for every $x$ in this subball $B'$ that $d(f_{i}(x),f_{j}(x))\leq\frac{1}{2N}$ whenever $i,j\geq M$, a rather striking conclusion given the definition of $E_{M}$ as an infinite double intersection. Now, holding $M$ fixed we find $$\lim\limits_{j\to\infty}d(f_{M}(x),f_{j}(x))=d(f_{M}(x),f(x))\leq\frac{1}{2N}.$$ An application of the triangle inequality then shows $B'\cap D_{N}=\emptyset$, and since the choice of $B$ was arbitrary (in particular, independent of $N$), we see that every ball $B$ in $X$ contains a subball $B'$ which avoids those points of discontinuity of $f$ lying in $D_{N}$ (i.e. those points where $\omega_{f}(x)>\frac{1}{N}$). Therefore, in sending $N\to\infty$, we conclude every ball $B$ of $X$ actually contains a subball $B'$ on which $f$ is continuous; moreover, we see that $D$ is the union of nowhere dense sets $D_{n}$, and so $f$ is in fact continuous on a dense set of $X$!
Problem #8.
 See Exam Handout.
Solution.

The RieszFischer representation theorem (Theorem 1.10.11) implies the existence of a unique positive Radon measure $\mu$ such that $I$ is represented by $\mu$, in the sense that
$$I(f)=\int_{\mathbb{R}}f\;d\mu$$
for every $f\in C_{c}(\mathbb{R}\to\mathbb{R})$ (obviously $\mathbb{R}$ is a locally/$\sigma$ compact Hausdorff space). But $I$ is translation invariant in and of itself, so the induced measure $\mu$ must also be translation invariant in the sense that
$$I(f_{h})=\int_{\mathbb{R}}f(xh)\;d\mu(x)=\int_{\mathbb{R}}f(x)\;d\mu(xh)=\int_{\mathbb{R}}f(x)\;d\mu(x)=I(f).$$
Thus, for any $E\subset\mathbb{R}$ and $x\in\mathbb{R}$, we must have
$$\mu(x+E)=\mu(E).$$ But then $\mu$ is just a multiple of the Lebesgue measure $m$ since by Exercise 1.2.23 in the previous textbook on measure theory, $\mu$ satisfies (i), (ii), and (iii) and the constant $c$ corresponds to the normalization of $\mu$, that is $\mu([0,1])=c$ and so
$$I(f)=\int_{\mathbb{R}}f\;d\mu=c\int_{\mathbb{R}}f\;dx$$
as claimed, where $dx=dm(x)$, the Lebesgue measure on $\mathbb{R}.$ The above arguments are a little symbolic and not very rigorous, so let us be explicit. For $I(f)=I(f_{h})$ to hold, we must have
$$\int_{\mathbb{R}}f(x+h)\;d\mu=\lim_{n\to\infty}\int_{\mathbb{R}}\phi_{n}(x+h)\;d\mu=\lim_{n\to\infty}\int_{\mathbb{R}}\phi_{n}(x)\;d\mu=\int_{\mathbb{R}}f(x)\;d\mu$$
for a monotonic sequence of simple functions converging to $f$ a.e. pointwise (such a sequence exists, e.g. one can use the standard construction of partitioning the range of $f$ and pulling back to measurable sets in order to define each $\phi_{n}$, and the result is implied by the monotone convergence theorem). But the integral of each $\phi_{n}$ is just
$$\int_{\mathbb{R}}\phi_{n}(x)\;d\mu=\sum\limits_{i=1}^{n}\alpha_{j}\mu(E_{j})$$
for measurable sets $E_{j}$ and scalars $\alpha_{j}$ (one can arrange the $E_{j}$ to be pairwise disjoint), and so
$$\int_{\mathbb{R}}\phi_{n}(xh)\;d\mu=\sum\limits_{i=1}^{n}\alpha_{j}\mu(E_{j}h).$$ Thus we see that the translation invariance of the integral forces translation invariance of the measure $\mu$. Now we show $\mu(E)=cm(E)$ for every $E\subset\mathbb{R}$ that is Lebesgue measurable. As usual, set $c=\mu([0,1])$ and note that $m([0,1])=1.$ Now consider the $n$ dyadic mesh on $\mathbb{R}$ consisting of $2^{n}$ disjoint intervals $I$ of length $2^{n}$. We know that the Lebesgue measure $m$ assigns a mass of $2^{n}$ to each such interval, and that the set $[0,1]$ can be assimilated by a union $2^{n}$ intervals of length $2^{n}$, all of which are translates of each other. The translation invariance of both $\mu$ and $m$ then implies for every $I$ in the $n$th dyadic mesh (e.g. $I=2^{n}$) that
$$c=\mu([0,1])=\sum_{j=1}^{2^{n}}\mu(I+h_{j})=\sum_{j=1}^{2^{n}}\mu(I)=2^{n}\mu(I)=cm([0,1])=c2^{n}m(I)=cm([0,1])$$
In other words,
$$\mu(I)=cm(I)$$
for every dyadic box of size $2^{n}$. Standard arguments from measure theory then implies $m$ and $\mu$ agree on all Borel sets (in particular, because every open set is the countable union of dyadic boxes), and thus $m$ and $\mu$ agree on the Lebesgue $\sigma$algebra since every such set disagrees with a Borel set on a null set. Passing from this measure to its associated integral, we then conclude
$$I(f)=\int_{\mathbb{R}}f\;d\mu=c\int_{\mathbb{R}}f\;dx$$
as required.
Exercise 1 (a) 1.1.3, (b) 1.1.4, (c) 1.1.7
 (Uniqueness of elementary measure). Let $d\geq1$, $m':\mathscr{E}(\mathbb{R}^{d})\to\mathbb{R}^{+}$ be a nonnegative, finite additive and translationinvariant map from the elementary sets to the positive reals. Prove the existence of a positive constant $c$ such that $m'(E)=cm(E)$ for all $E\in\mathscr{E}(\mathbb{R}^{d})$.
 (Products of elementary sets are elementary). Let $d_{1},d_{2}\geq1$, $E_{1}\subset\mathbb{R}^{d_{1}}$ and $E_{2}\subset\mathbb{R}^{d_{2}}$ elementary, and show that $E_{1}\times E_{2}\subset\mathbb{R}^{d_{1}+d_{2}}$ is elementary with $m(E_{1}\times E_{2})=m(E_{1})m(E_{2})$.
 (Regions under graphs of continuous functions are Jordan measurable). Let $B\subset\mathbb{R}^{d}$ be a closed box and $f\in\mathscr{C}[B\to\mathbb{R}^{d}]$. (1) Show that the graph $G_{f}=\{(x,f(x))\;:\;x\in B\}\subset\mathbb{R}^{d+1}$ has Jordan measure zero. (2) Show that the set $\mathscr{A}=\{(x,t)\;:\;x\in B,\;0\leq t\leq f(x)\}\subset\mathbb{R}^{d+1}$ is Jordan measurable.
Solution.
 Let $m$ be the usual elementary measure, $m'$ as above, and put $c=m'\left([0,1)\right)$. Then
\begin{align*}
c=m'([0,1))
&=m'\left(\bigcup\limits_{j=0}^{n}\left[\frac{j}{n},{j+1}{n}\right)\right) &\text{(finite union of disjoint elementary sets)}\\
&=\sum\limits_{j=1}^{n}m'\left(\left[\frac{j}{n},\frac{j+1}{n}\right)\right) &\text{(finite additivity)}\\
&=\sum\limits_{j=1}^{n}m'\left(\left[0,\frac{1}{n}\right)\right) &\text{(translation invariance)}\\
&=n\cdot m'\left[0,\frac{1}{n}\right)\\
\end{align*}
so that
$$m'\left(\left[0,\frac{1}{n}\right)\right)=\frac{c}{n}.$$
Therefore $m'$ and $m$ agree on intervals of the form $[0,\frac{1}{n})$ upto a constant normalization of $[0,1)$; in particular, if $c=1$, they agree exactly on these intervals, and so in the following we may use the phrase "agree" without reference to normalization. Repeated applications of translation invariance and finite additivity now shows that $m$ and $m'$ also agree on all intervals $[a,b)$ with rational end points. Restricting attention to such sets, and treating $m'$ as a function of the end points, we have that
\begin{align*}
\Bigm'(\vec{x_{1}x_{2}})\Big
&= \Bigm'\left([a_{1},b_{1})\right)m'\left([a_{2},b_{2})\right)\Big\\
&= c\Bigm\left([a_{1},b_{1})\right)m\left([a_{2},b_{2})\right)\Big \\
&= c\Big(b_{1}a_{1})(b_{2}a_{2})\Big \\
&= c\Big(b_{1}b_{2})(a_{1}a_{2})\Big \\
&\leq c\Bigb_{1}b_{2}+a_{1}a_{2}\Big\\
&\leq c\Bigx_{1}x_{2}\Big, &\text{(equivalent $\ell_{\infty}$ norm)}
\end{align*}
which shows that $m'$ (and $m$) is Lipschitz continuous with Lipschitz constant $c$ ($1$) on the set of intervals $[a,b)$ with rational end points; and since $\mathbb{Q}$ is dense in $\mathbb{R}$, it follows that $m$ and $m'$ agree on all intervals $[x,y)$ with real end points. Applying finiteadditivity again then shows $m$ and $m'$ agree on sets of the form $\bigcup_{k=1}^{N}[x_{k},y_{k})$, which are the elementary sets of $\mathbb{R}^{1}$. It follows that $m$ and $m'$ agree for all elementary sets in $\mathbb{R}^{d}$ with normalization constant $c\mapsto c^{d}$.
 We have $E_{1}=\bigcup_{j=1}^{N} B^{1}_{j}$ and $E_{2}=\bigcup_{j=1}^{M} B^{2}_{j}$. Therefore
\begin{align*}
E_{1}\times E_{2}
&= \left(B^{1}_{1}\cup\ldots\cup B^{1}_{N}\right)\times\left(B^{2}_{1}\cup\ldots\cup B^{2}_{M}\right)\\
&=\Big(\{B^{1}_{1}\times B^{2}_{1}\}\cup\ldots\cup \{B^{1}_{1}\times B^{2}_{M}\}\Big)\cup\ldots\cup\Big(\{B^{N}_{1}\times B^{2}_{1}\}\cup \ldots\cup \{B^{N}_{1}\times B^{2}_{M}\}\Big)\\
&= \bigcup\limits_{j=1}^{N}\bigcup\limits_{i=1}^{M}B^{1}_{j}\times B^{2}_{i} \\
&= \bigcup_{k=1}^{NM}B'_{k}
\end{align*}
where $k$ indexes each cross product (which is again a box $B'_{k}$). Since $NM<\infty$, we see that $E_{1}\times E_{2}$ is indeed elementary. Furthermore, by part $(i)$ of Lemma $1.1.2$ from the text, we may assume that the sets $\{B^{1}_{j}\}_{j=1}^{N}$ and $\{B^{2}_{j}\}_{j=1}^{M}$ are pairwise disjoint so that the set $\bigcup_{ji}\{B^{1}_{j}\times B^{2}_{i}\}=\{B'_{k}\}_{k=1}^{NM}$ is also pairwise disjoint. Moreover, by part $(ii)$ we also have $m(E_{1})=\sum_{j=1}^{N}B^{1}_{j}$ and $m(E_{2})=\sum_{j=1}^{M}B^{2}_{j}$, so that
\begin{align*}
m(E_{1}\times E_{2})
&= m\left(\bigcup\limits_{k=1}^{NM}B'_{k}\right) \\
&= \sum\limits_{k=1}^{NM}B'_{k}\\
&= \sum\limits_{j=1}^{N}\sum\limits_{i=1}^{M}B^{1}_{j}\times B^{2}_{i} \\
&= \sum\limits_{j=1}^{N}\sum\limits_{i=1}^{M}B^{1}_{j}B^{2}_{i}\\
&= \sum\limits_{j=1}^{N}B^{1}_{j}\sum\limits_{i=1}^{M}B^{2}_{i}\\
&= m(E_{1})m(E_{2})
\end{align*}
as required.
 For this part, let $\mathscr{G}_{f}$ denote the graph of $f$ and $\mathscr{R}^{+}_{f}$ denote the region of $f$ between the coordinate plane and its positive values.
To prove (1) let $\epsilon>0$ be given. The compactness of $B$ implies the existence of a $\delta_{\epsilon}$ (depending on $\epsilon$ only) such that $xy<\delta_{\epsilon}$ implies $f(x)f(y)<\epsilon$ for all $x,y\in B$. Corresponding to this $\delta_{\epsilon}$, put $\delta\leq\frac{1}{\sqrt{d}}\cdot\delta_{\epsilon}$. Then clearly $\delta$ also satisfies the uniform continuity condition; moreover, if $Q$ has center $x$ and $Q=\delta\times\ldots\times\delta$, then $Q\subset B(x;\delta_{\epsilon})$. Now suppose $B=\ell_{1}\times\ldots\times\ell_{d}$, and define integers $\{N_{i}\}_{i=1}^{d}$ by $$N_{i}=\left\lceil\frac{\ell_{1}}{\delta}\right\rceil=min\{n\in\mathbb{N}\;:\;n\geq\frac{\ell_{1}}{\delta}\}.$$ Then $B$ can be covered by a family of $M=\prod_{i=1}^{d} N_{i}<\infty$ pairwise disjoint cubes $\{Q_{k}\}_{k=1}^{M}\subset\mathbb{R}^{d}$, each of dimension $\ell=\delta$, hence volume $\delta^{d}$. This is achieved in the obvious way by first extending each $\ell_{i}$ of $B$ by $N_{i}\delta\ell_{i}$, then partitioning the resulting lengths into $N_{i}$ disjoint subintervals of length $\delta$, and then finally by taking the union of $M$ cross products in a manner similar to part $(b)$. From this cover of $B$, we construct a family of of $M$ pairwise disjoint boxes $\{B_{k}\}_{k=1}^{M}\subset\mathbb{R}^{d+1}$ by defining $B_{k}=Q_{k}\times I^{\epsilon}_{k}$, where $I^{\epsilon}_{k}=[\inf_{x\in Q_{k}}f(x),\sup_{x\in Q_{k}} f(x)]$, so that by uniform continuity $I^{\epsilon}_{k}<\epsilon$. Then $$G_{f}\subset\bigcup_{k=1}^{M}B_{k},$$
and we have
\begin{align*}
m^{\star,(J)}(G_{f})
&=\inf\limits_{B_{1}\cup\ldots\cup B_{N}\supset G_{f}}\sum\limits_{j=1}^{N}B_{j} \\
&\leq\sum\limits_{k=1}^{M}B_{k}\\
&=\delta^{d}\sum\limits_{k=1}^{M}\sup\limits_{x\in Q_{k}}f(x)f(y)\\
&\leq M\delta^{d}\epsilon.
\end{align*}
Since $\epsilon$ is arbitrary, and $O(M\delta^{d}\epsilon)=O(\prod_{i=1}^{d}N_{i}\cdot\delta^{d}\epsilon)=O(\delta^{d}\delta^{d}\epsilon)=O(\epsilon)$, we see that $M\delta^{d}\epsilon\to0$ as $\epsilon\to0$. Hence, $m^{\star,(J)}(G_{f})=0$ which implies that $m(J)(G_{f})=0$.
To prove (2) we make a slight modification to the construction of the $Q_{k}$ above, resulting in a family of cubes and boxes instead. We now keep each $\ell_{i}$ fixed and redefine $$N_{i}=\left\lfloor\frac{\ell_{1}}{\delta}\right\rfloor=max\{n\in\mathbb{N}\;:\;n\leq\frac{\ell_{1}}{\delta}\}.$$ Then each $\ell_{i}$ can be partitioned into $N_{i}$ subintervals of length $\delta$ in addition to a ``fringe'' interval $F_{i}$ of length at most $\delta$. We now construct a family of pairwise disjoint boxes $\{Q_{k}\}_{k=1}^{M}$ ($M=O(\delta^{d})<\infty$) which cover $B$ by taking the union of all cross products from the resulting subinterval partitions. Note that any $Q_{k}$ arising from poducts involving some $F_{i}$ have volume $Q_{k}<\delta^{d}$, and in fact, because of the way the partition is made, $B=\bigcup_{k=1}^{M}Q_{k}$. Now assume $f$ is nonnegative (so that $\mathscr{A}$ contains the entire region under $f$) and define numbers $\{\alpha_{k}\}_{k=1}^{M}$ by $\alpha_{k}=\sup_{x\in Q_{k}}f(x)$ (this exists and is finite because of compactness). Then $\alpha_{k}f(x)<\epsilon$ for every $x\in Q_{k}$ (by uniform continuity). We can then construct a finite sequence of boxes $\{B^{\alpha}_{k}\}_{k=1}^{M}$, each of volume $\alpha_{k}Q_{k}$, which cover the set $\mathscr{A}$ by setting $B_{k}=Q_{k}\times[0,\alpha_{k}]$. Likewise, if we define numbers $\{\beta_{k}\}_{k=1}^{M}$ by $\beta_{k}=\inf_{x\in Q_{k}}f(x)$, we obtain in the same manner a finite sequence of boxes $\{B^{\beta}_{k}\}_{k=1}^{M}$ of volume $\beta_{k}Q_{k}$ which are contained in $\mathscr{A}$. (Essentially, we have formed upper and lower Darboux sums of $f$ on $B$). We now take differences to obtain \begin{align*} \left\sum\limits_{k=1}^{M}B^{\alpha}_{k}\sum\limits_{k=1}^{M}B^{\beta}_{k}\right &= \sum\limits_{k=1}^{M}B^{\alpha}_{k}B^{\beta}_{k}\\ &=\sum\limits_{k=1}^{M}Q_{k}\alpha_{k}\beta_{k}\\ &\leq\delta^{d}\sum\limits_{k=1}^{M}\alpha_{k}\beta_{k}\\ &<\delta^{d}M\epsilon. \end{align*} Recalling that $\delta^{d}M=O(1)$ (even with the modified construction) and appealing to the obvious inequality $$\sum_{k}^{M}B^{\beta}_{k}\leq m_{\star,(J)}(\mathscr{A})\leq m(\mathscr{A})\leq m^{\star,(J)}(\mathscr{A})\leq\sum_{k}^{M}B^{\alpha}_{k},$$ we see that $$m^{\star,(J)}(\mathscr{A})m_{\star,(J)}(\mathscr{A})\leq\epsilon$$ so that $\mathscr{A}$ is Jordan measurable. The case where $f$ is possibly negative now follows immedaitely by observing that $m(\mathscr{A}_{f})=m(\mathscr{A}_{g})$ where $g$ is the function defined by $g(x)=f(x)$ for $x$ such that $f(x)\geq0$ and $g(x)=0$ for $x$ such that $f(x)<0 databloggerescapedabove="" databloggerescapedand="" databloggerescapedapplies="" databloggerescapedcontinuity="" databloggerescapedcontinuous="" databloggerescapedf="" databloggerescapedfunction="" databloggerescapedg="" databloggerescapedimplies="" databloggerescapedli="" databloggerescapednonnegative="" databloggerescapednote="" databloggerescapedof="" databloggerescapedproof="" databloggerescapedso="" databloggerescapedthat="" databloggerescapedthe="" databloggerescapedto="">
Exercise 1 (a) 1.1.8, (b) 1.1.9, (c) 1.1.10
 (Measurability of triangular shapes). Suppose $A,B,C\in\mathbb{R}^{2}$. (1) Show that the solid triangle with vertices $A,B,C$ is Jordan measurable. (2) Show that the Jordan measure of the solid triangle is equal to $\frac{1}{2}(BA)^(CA)$ where $(a,b)^(c,d):=adbc$.
 (Measurability of compact convex polytopes). Show that every compact convex polytope in $\mathbb{R}^{d}$ is Jordan measurable.
Solution.
 (1) It is obvious that $\mathscr{T}$ is expressible as the region $\mathscr{R}_{f\geq g}$ bounded by two (possibly piecewise) continuous linear functions defined on the projection of $\mathscr{T}$ onto the coordinate $x$ axis, and so we may apply corollary 1 to conclude $\mathscr{T}$ is Jordan measurable. Note that it does not matter if $\mathscr{T}$ has an edge parallel to any of the coordinate axes. Note also that we can now use induction to prove any $d$ dimensional triangle is Jordan measurable. Indeed, having established the base case, suppose any $(d1)$ dimensional triangle is Jordan measurable. Then any $d$ dimensional triangle is bounded between two continuous functions defined on the projection of the $d$ dimensional triangle onto the $(d1)$ dimensional coordinate axes, and this is either a $(d1)$ dimensional triangle or a box. Hence, corollary 1 shows that the $d$ dimensional triangle is Jordan measurable.
(2) To get an idea of what the given expression is, we expand its definition: \begin{align*} (BA)\hat{\;\;\;}(CA) &= \frac{1}{2}((B_{x}A_{x},B_{y}A_{y})^(C_{x}A_{x},C_{x}C_{y})\\ &= \frac{1}{2}(B_{x}A_{x})(C_{y}A_{y})(B_{y}A_{y})(C_{x}A_{x})\\ &= \frac{1}{2}B_{x}C_{y}+B_{y}A_{x}+A_{y}C_{x}B_{x}A_{y}A{x}C_{y}B_{y}C_{x}\\ &= \frac{1}{2}A_{x}(B_{y}C_{y})B_{x}(A_{y}C_{y})+C_{x}(A_{y}B_{y}). \end{align*} It is clear now (and in retrospect, even before the computation) that this formula is equivalent to the $\vec{k}$ minor associated with the cross product determinant of $\vec{BA}\times\vec{CA}$, and because the $\vec{i}$ and $\vec{j}$ minor determinants are both $0$ (since the vectors spanning $\mathscr{T}$ are planar), it is also equal to the magnitude $\vec{BA}\times\vec{CA}$. Therefore, if $\theta$ is the angle between the vectors $\vec{AB}$ and $\vec{AC}$, we have that \begin{align*} \frac{1}{2}(BA)\hat{\;\;\;}(CA) &=\vec{AB}\times\vec{AC}\cdot\vec{k}\\ &=\vec{AB}\times\vec{AC}\\ &=\frac{1}{2}\vec{AB}\vec{AC}\sin\theta\\ &=\frac{1}{2}Base(\mathscr{T})Height(\mathscr{T}), \end{align*} which is the well known formula for the area (measure) of $\mathscr{T}$.  From the definition, any compact convex polytope $P$ is the intersection of $N$ half spaces $H_{n}$, and so can be written as the solution to a system of linear inequalities: $$Ax\leq c,$$ or more explicitly $$\sum_{j=1}^{d}x_{i}v_{i}\leq c_{k},\;\;k=1,2,\ldots,N.$$ The solution to such a system of inequalities is a region bounded by a set of planes, and this case, the region is assumed to be bounded. The result is trivial if $d=1$, since the only half spaces are intervals, the finite intersection of which clearly being Jordan measurable. For $d=2$, it is again trivial, for the projection of $P$ onto the $x$axis is just a closed interval, and we can apply corollary 1 since any half space can be written as a continuous linear function, and therefore $P$ can be expressed as the region $\mathscr{R}_{f\geq g}$ bounded by two (possibly piecewise) continuous linear functions defined on this interval. We now apply induction for the general case. Indeed, having established the base case, suppose any $(d1)$ dimensional compact convex polytope is Jordan measurable. Then any $d$ dimensional compact convex polytope is bounded between two continuous functions defined on the projection of the $d$ dimensional compact convex polytope onto the $(d1)$ dimensional coordinate axes, and this in turn is a $(d1)$ dimensional compact convex polytope. Hence, corollary 1 shows that the $d$ dimensional compact convex polytope is Jordan measurable.
 (1) We first handle the case of a closed ball $B$, and the argument is identical to the previous parts of this exercise. If $d=1$, then any closed ball $B$ is just a closed interval, hence Jordan measurable. If $d=2$, then the projection of $B$ onto the $x$axis is itself an interval, and corollary 1 applies since $B$ may be expressed as the region $\mathscr{R}_{f\geq g}$ of two continuous (square root) functions defined on this interval. Induction then shows that any $d$dimensional closed ball is Jordan measurable. Indeed, having established the base case, assume this. Then any $d$ dimensional closed ball is bounded between two continuous functions defined on the projection of the $d$ dimensional closed ball onto the $(d1)$ dimensional coordinate axes, and this in turn is a $(d1)$ dimensional closed ball. Hence, corollary 1 shows that the $d$ dimensional compact closed ball is Jordan measurable. To handle the case of an open ball $B^{\circ}$, we note that $\partial B^{\circ}=\mathscr{G}_{f}\cup\mathscr{G}_{g}$, and therefore $m^{\star,(J)}(\partial B^{\circ})=0$. Consequently, $m^{\star,(J)}(\bar{B}\Delta B^{\circ})=0$, so that $B^{\circ}$ is Jordan measurable since we just proved $\bar{B}$ is. Furthermore, their measures coincide since we may now apply additivity to obtain $m(\bar{B})=m(B^{\circ}\cup\partial B)=m(B^{\circ})+m(\partial B)=m(B^{\circ}).$
To prove that $m(B_{r})=c_{d}r^{d}$, consider the open unit ball (taking the closure if necessary) centered at the origin \begin{align*} B_{1}(0) &=\left\{x\in\mathbb{R}^{d}:x<1 databloggerescaped1="" databloggerescaped2.="" databloggerescaped2="" databloggerescapeda="" databloggerescapedalign="" databloggerescapedan="" databloggerescapedand="" databloggerescapedany="" databloggerescapedapplying="" databloggerescapedare="" databloggerescapedat="" databloggerescapedb="" databloggerescapedb_="" databloggerescapedball="" databloggerescapedbe="" databloggerescapedbegin="" databloggerescapedbeginning="" databloggerescapedbehaves="" databloggerescapedbelow="" databloggerescapedbigcup="" databloggerescapedby="" databloggerescapedc_="" databloggerescapedcenter="" databloggerescapedcompleteness.="" databloggerescapedcomponent="" databloggerescapedcompute="" databloggerescapedcomputed="" databloggerescapedconsequence="" databloggerescapedcorollary="" databloggerescapedcould="" databloggerescapedcovers="" databloggerescapedcrucial="" databloggerescapedcup="" databloggerescapedd="" databloggerescapeddelta="" databloggerescapeddelta_="" databloggerescapeddilating="" databloggerescapeddilation="" databloggerescapeddilations="" databloggerescapeddirectly="" databloggerescapeddisjoint="" databloggerescapeddistributive="" databloggerescapede.g.="" databloggerescapede="" databloggerescapedeach="" databloggerescapedeasy="" databloggerescapedelementary="" databloggerescapedend="" databloggerescapedequal="" databloggerescapedevery="" databloggerescapedexercise="" databloggerescapedextends="" databloggerescapedfact="" databloggerescapedfactor="" databloggerescapedfollows="" databloggerescapedfor="" databloggerescapedfrom="" databloggerescapedgives="" databloggerescapedh="" databloggerescapedhand="" databloggerescapedhas="" databloggerescapedhave="" databloggerescapedhere="" databloggerescapedhow="" databloggerescapedi="1}^{d}\delta_{i}=r^{d}$." databloggerescapedidea="" databloggerescapedif="" databloggerescapedimmediately="" databloggerescapedin="" databloggerescapedincidentally="" databloggerescapedinf_="" databloggerescapedinfimum="" databloggerescapedinvariance="" databloggerescapedinvariant="" databloggerescapedis="" databloggerescapedisn="" databloggerescapedit="" databloggerescapedj="" databloggerescapedjordan="" databloggerescapedknow="" databloggerescapedlast="" databloggerescapedldots="" databloggerescapedleft="" databloggerescapedlet="" databloggerescapedli="" databloggerescapedlimits_="" databloggerescapedm="" databloggerescapedmathbb="" databloggerescapedmeasurable="" databloggerescapedmeasure="" databloggerescapedmoreover="" databloggerescapedno="" databloggerescapednote="" databloggerescapednow="" databloggerescapedobservations="" databloggerescapedobtain="" databloggerescapedobtained="" databloggerescapedobvious="" databloggerescapedof="" databloggerescapedon="" databloggerescapedother="" databloggerescapedover="" databloggerescapedplausible="" databloggerescapedpossible="" databloggerescapedpresent="" databloggerescapedprod="" databloggerescapedproperty="" databloggerescapedprove="" databloggerescapedr="" databloggerescapedrb_="" databloggerescapedre="" databloggerescapedreally="" databloggerescapedreduces="" databloggerescapedrelatively="" databloggerescapedresult="" databloggerescapedright="" databloggerescapedrx:x="" databloggerescapedrx_="" databloggerescapedsake="" databloggerescapedsee="" databloggerescapedset.="" databloggerescapedsets="" databloggerescapedsince="" databloggerescapedso="" databloggerescapedsome="" databloggerescapedsubsequent="" databloggerescapedsubset="" databloggerescapedsuch="" databloggerescapedsum="" databloggerescapedsuppose="" databloggerescapedsupset="" databloggerescapedt="" databloggerescapedtext="" databloggerescapedthat="" databloggerescapedthe="" databloggerescapedthen="" databloggerescapedthis="" databloggerescapedthough="" databloggerescapedto="" databloggerescapedtranslating="" databloggerescapedtranslation="" databloggerescapedtrue="" databloggerescapedunder="" databloggerescapedunderstood="" databloggerescapeduniform="" databloggerescapeduniformly="" databloggerescapedunit="" databloggerescapeduseful="" databloggerescapedvery="" databloggerescapedway="" databloggerescapedwe="" databloggerescapedwhere="" databloggerescapedwhich="" databloggerescapedwill="" databloggerescapedxh:x="" databloggerescapedxh="" databloggerescapedx="" databloggerescapedx_="" databloggerescapedyields="">
(2) Crude lower and upper bounds for $c_{d}$ can be established by computing the measure of the supremum and infimum covers of $B_{1}(0)$ by a single cube from inside and outside, respectively, for $d=1,2,\ldots$. To begin, let's examine a few concrete cases. When $d=1$ there is no need to compute bounds since $m^{1}(B_{1}(0))=2$. For $d=2$, it is clear that the supremal cube ($E_{2}$) has a diagonal length of $\mathscr{D}=2$ and that the infimal cube ($F_{2}$) has a dimensional length of $\ell=2$ (this can be proved easily using simple optimization techniques from differential calculus). An application of the Pythagorean theorem then shows that $\mathscr{D}^{2}=2\ell^{2}$ for any cube in $\mathbb{R}^{2}$, hence $m(E_{2})=2$ and $m(F_{2})=4$. When $d=3$, we proceed similarly. The infimal cube $F_{3}$ has again dimensional length $\ell=2$ and hence $m(F_{3})=2^{d}$. Likewise, the supremal cube has diagonal length $\mathscr{D}=2$. An application of the Pythagorean theorem does not proceed as simply for $d>2$. In any case, we have $\mathscr{D}^{2}=\ell^{2}+\mathscr{D}^{'2}$ where $\mathscr{D}^{'}$ is the projection of $\mathscr{D}$ onto $\mathbb{R}^{2}$ (a picture is helpful for the cases $d=2,3$). Then $\mathscr{D}^{'}$ forms the diagonal of a righttriangle in $\mathbb{R}^{2}$ with sides $\ell$. Hence $\mathscr{D}^{'}=\sqrt{2}\ell$, which implies $3\ell^{2}=\mathscr{D}^{2}$, so that $\ell=\frac{2}{\sqrt{3}}$, thus obtaining $m(E_{3})=\frac{8}{3\sqrt{3}}$. Therefore we assert that $$\left(\frac{2}{\sqrt{d}}\right)^{d}\leq c_{d}\leq2^{d},$$ which holds for (so far) $d=1,2,3$, equality holding only for $d=1$. Now let's consider the situation for any $d+1$ (when I originally did this problem, I assumed induction would be necessary (hence the $d+1$), but it turns out not to be). The unit ball is certainly covered by a cube of dimensional length $\ell=2$ ($F_{d+1}$), and it covers a cube of diagonal length $\ell=2$ ($E_{d+1}$). (Note that we are no longer asserting such sets are infimal or supremal, but this doesn't matter at this stage.) This follows from the definition of $B_{1}(0)$ for any dimension since $x< 1$ implies $\left(\sum_{i=1}^{d+1}x_{i}^{2}\right)^{\frac{1}{2}}< 1$, and any point which satisfies this inequality is obviously in $F_{d+1}$ so that it covers $B_{1}(0)$. Likewise, the point $x=(1,0,\ldots,0)$ is clearly not in $E_{d+1}$, so that $B_{1}(0)$ covers $E_{d+1}$. Since all sets in question are measurable, we have that $$m(E_{d+1})\leq m(B_{1}(0))\leq m(F_{d+1}).$$ Therefore, all we must do is compute the measures of $E_{d+1}$ and $F_{d+1}$, and if they agree with the conjectured bounds, then the bounding inequality (because of the inequality just proved) holds for all $d=1,2,\ldots$. Clearly we have $m(F_{d+1})=2^{d+1}$ by definition. As for $E_{d+1}$ we take the diagonal $\mathscr{D}=2$ passing through the origin and form a right triangle by extending one line directly down the length of the cube at the point of intersection with $B_{1}(0)$. This line is the component orthogonal to the orthogonal projection $\mathscr{D}^{'}$ of $\mathscr{D}\subset\mathbb{R}^{d+1}$ into $\mathbb{R}^{d}$. Now $\mathscr{D}^{'}$ is the diagonal emanating through the origin to the intersection(s) of the projection of $B_{1}$ and $E_{d+1}$ into $\mathbb{R}^{d}$. Hence the "vertical" side is the new projection line of $\mathscr{D}^{'}\mapsto\mathbb{R}^{d2}$ of length $\ell$ and the ``horizontal line'' is $\mathscr{D}^{'''}$. This process continues until $\mathscr{D}^{d}=\sqrt{2}\ell$ is obtained. Then backsubstituting and applying the Pythagorean theorem at each stage yields $\mathscr{D}^{2}=4=\ell^{2}+d\ell^{2}$ which implies $\ell=\frac{2}{\sqrt{d+1}}$. Hence $$m(E_{d+1})=\left(\frac{2}{\sqrt{d+1}}\right)^{d+1}$$ as required.
It can evidently be proven that the cubes $E_{d}$ and $F_{d}$ are indeed the supremal contained and infimal covering single cubes for $B_{1}(0)$ in $\mathbb{R}^{d}$. For general radius $B_{\delta}(0)$, we see that in the above formulas $2\mapsto 2\delta$. Therefore, the respective cubes are $E_{d}=\left(\frac{2\delta}{\sqrt{d}}\right)^{d}$ and $F_{d}=(2\delta)^{d}$. These cubes are useful in various instances.
Exercise 3. (a) 1.1.11
 (Relative invariance of Jordan measure under linear operators). Let $L:\mathbb{R}^{d}\to\mathbb{R}^{d}$ be a linear operator. (1) Show that there exists a real number $D$ such that $m(L(E))=Dm(E)$ for every elementary set $E$. (2) Show that if $E$ is Jordan measurable, then $L(E)$ is also, and that $m(L(E))=Dm(E)$. (3) Show that $D=det L$.
Solution.
 We first record some quick facts about how $L$ affects a general set $S$. A general linear operator on a finite $d$dimensional real vector space can always be represented by a $d\times d$ real matrix $[L]_{ij}$ (the standard basis is assumed). Furthermore, a sequence of elementary row operations always exists such that either $[L]$ is reduced to the identity matrix $[I]$ ($L$ is an isomorphism) or to $[I]$ with some finite number of null rows ($L$ is singular). Call such row operations ``elementary operators'' $e:\mathbb{R}^{d}\to\mathbb{R}^{d}$ represented by matrices $[e]$, which are obtained by applying the corresponding single row operation to $[I]$. Since each elementary operator is invertible (since the corresponding row operations are), it is evident that if we record the sequence of row operations that bring $[L]\to[I]$, we can reobtain $[L]$ from $[I]$ by applying the inverse row operations in reverse sequence to $[I]$. Therefore, we see that $[I]=[L][e_{1}]...[e_{n}]$ and $[L]=[I][e_{n}]^{1}...[e_{1}]^{1}$, or in other words $$L(x)=(e_{n}^{1}\circ e_{n1}^{1}\circ\ldots\circ e_{1}^{1})(x).$$ (Although we have tacitly assumed $L$ is invertible, it is obvious that a singular $L$ still has a finite elementary operator decomposition). This means that if $L$ is invertible, then $L(S)$ is a sequence of coordinate plane reflections (row interchanges), proper shears (linear combinations of independent rows), and dilations (nonzero scalar multiples of rows) of $S$. Note that when dilation is by a negative scalar (in particular, $1$), $S$ is reflected along the corresponding coordinate axis. Note also that a rigid motion (in particular, a rotation) is accomplished by a combination of reflections, shears and dilations. If $L$ is singular, then in addition to the above, $L(S)$ necessarily has a subsequence of improper (e.g. projective) shears (linear combinations of dependent rows) or projections onto coordinate subspaces (coordinate dilations by $0$). Note that if $L$ is affine (e.g. $L=T+b$), then a translation of $S$ is also possible, and if we define $\det T=\det L$, then this exercise applies to affine transformations as well. However, we shall not discuss affine transformations further, since our proof requires the translation invariance of $m$, and this is already easy to prove by itself.
(1) Let $E\in\mathscr{E}\left(\mathbb{R}^{d}\right)$ and first assume $L$ is invertible with $d\geq2$. Then $L(E)$ is either a dilation of $E$, a reflection of $E$ about a coordinate plane, a reflection of $E$ about a coordinate axis, a shear of $E$ about some fixed axis, or some combination of these. If $L$ is a dilation of $E$ or a reflection of $E$ about a coordinate axis, then clearly $L(E)$ is still an elementary set, so it is Jordan measurable by definition. Likewise, if $L$ is a reflection of $E$ about a coordinate plane, then because such a reflection amounts to a 90 degree rotation, we see that $E$ is again an elementary set. To deal with the final possibility of a shear, first consider a single box $B\subset\mathbb{R}^{d}$. If $L(B)$ is shear of $B$, then it is a $d$dimensional parallelogram. Since any parallelogram is the union of two $d$dimensional triangles (e.g. $d$simplexes), the $d$dimensional equivalent of exercise 1.1.8 applies and we conclude $L(B)$ is Jordan measurable (the modification is straight forward since it follows from corollary 1 and the fact that a $d$dimensional triangle can be written as the region bounded by two piecewise continuous linear functions $f,g:\mathbb{R}^{d1}\to\mathbb{R}$). The general case $L$ now follows, since $E$ is the finite union of boxes, $L(E)$ is the finite union of (possibly translated, dilated, and/or reflected) triangles, all of which are Jordan measurable. Hence $L(E)$ is Jordan measurable because Jordan measurable sets are closed under finite unions. If $L$ is singular, then let $d'=\ker L\geq1$. If $d'\geq2$, then $L(E)$ has outer Jordan measure $0$, since it is a subset of a $d2$ dimensional subspace, and any such set can be covered by degenerate $d1$ dimensional boxes, which all have $d$dimensional measure $0$. This means that its inner Jordan measure is also $0$, so $L(E)$ is Jordan measurable. If $d'=1$, then $L(E)$ is a subset of a $d1$ dimensional subspace, but in general (because of rotations) cannot be covered by $d1$ dimensional cubes. On the other hand, $L(E)$ is the graph of some continuous linear function $f:\mathbb{R}^{d1}\to\mathbb{R}$, and so it is Jordan measurable with measure $0$ by exercise 1.1.7. Finally, if $d=1$, then only a translation and or dilation of $E$ is possible (if $L$ is singular, then $L(S)=\{0\}$), which is evidently Jordan measurable.
Now that we know $L(E)$ is Jordan measurable, we can show $m(L(E))=Dm(E)$ for some constant $D$. Define the map $m'(E)=m(L(E))$, e.g. $E\mapsto m(L(E))$. Then clearly $m'$ is a nonnegative function from the elementary sets to $\mathbb{R}^{+}$, e.g. $\mathscr{E}(\mathbb{R}^{d})\mapsto \mathbb{R}^{+}$. Assume for now that $L$ is invertible. Let $E_{h}$ denote the translate of $E$ by $h$. It is obvious that because rotations, reflections, shears and dilations are all spatially invariant, that $L(E_{h}))=L(E)_{h}$. Combining this with the translation invariance of $m$ gives $$m'(E_{h})=m(L(E_{h}))=m(L(E)_{h})=m(L(E))=m'(E),$$ which shows $m'$ is also translation invariant. Since $L$ is invertible and $d<\infty$, $L$ is a bijective isomorphism (in particular, $L$ is onetoone), so that if $\{E_{i}\}_{i=1}^{N}$ is a finite collection of pairwise disjoint elementary sets, $\{L(E_{i})\}_{i=1}^{N}$ is a finite collection of pairwise disjoint Jordan measurable sets. Then, since $m$ is finitely additive, we have that \begin{align*} m'\left(\bigcup\limits_{i=1}^{N}E_{i}\right) &=m\left(L\left(\bigcup\limits_{i=1}^{N}E_{i}\right)\right)\\ &=m\left(\bigcup\limits_{i=1}^{N}L(E_{i})\right)\\ &=\sum\limits_{i=1}^{N}m(L(E_{i}))\\ &=\sum\limits_{i=1}^{N}m'(E_{i}), \end{align*} which shows $m'$ is also finitely additive. In the case that $L$ is singular, some of the $L(E_{i})$ may no longer be disjoint, so at present we only have subadditivity. The only way we can recover additivity is if all intersecting $L(E_{i})$ have measure $0$, and for this to hold requires $L$ to map any elementary set $E$ to a set of measure $0$. This is in fact the case, and is a consequence of the Rank Nullity theorem as discussed previously. Incidentally, this also proves $m'$ is translation invariant because $m'(E_{h})=m(L(E_{h}))=0=m(L(E)_{h})=m(L(E))=m'(E)$. Hence, for any $L$, $m'$ is nonnegative, translation invariant, and finitely additive, which implies $m'(E)=m(L(E))=Dm(E)$ by uniqueness upto normalization of elementary measure.
(2) Suppose $E$ is now a Jordan measurable set and assume $L$ is invertible (if $L$ is singular, then $L(E)$ has outer measure $0$, and is therefore Jordan measurable). Then for any $\epsilon>0$ there exists an elementary set $A$ such that $m^{\star,(J)}(A\Delta E)\leq\epsilon$. For convenience, let $F=A\Delta E$ and choose pairwise disjoint cubes $\{Q_{j}\}_{j=1}^{N}$ of lengths $\ell_{j}$ such that both $$F\subset\bigcup_{j=1}^{N}Q_{j}$$ and $$\sum\limits_{j=1}Q_{j}\leq m^{\star,(J)}(F)+\epsilon$$ hold for $\epsilon$ chosen. Now since $d<\infty$, $L$ is a bounded linear operator, and so it satisfies a Lipschitz condition on $\mathbb{R}^{d}$ for some $0
Note: The existence of the cover $\{Q_{j}\}_{j=1}^{N}$ follows from the definition of $m^{\star,(J)}$, since every bounded set of real numbers has an infimum. Cubes are chosen for convenience, but the Lipschitz estimate still works for arbitrary boxes (just more notation is required); that cubes can be chosen in the first place follows from the fact that any box can be arbitrarily approximated by a finite number of cubes through an obvious dissection process. The comparison criterion used at the end is proven in exercise 1.1.5 for comparison with elementary sets; but it is obvious the comparison is valid with Jordan measurable sets, since a Jordan measurable set is itself arbitrarily approximated by an elementary set.
(3) First of all, if $L$ is singular, then $m(L(E))=0$ as already shown, and since $\det L=0$, the theorem holds for this case. So now we can assume $L$ is invertible and consider its elementary operator decomposition (which always exists) $$L(x)=(e_{1}\circ e_{2}\circ\ldots\circ e_{n})(x),$$ or in matrix multiplication form $$[L]_{ij}=[e_{1}][e_{2}]\ldots[e_{n}].$$ Since the determinant of an operator is defined by the determinant of the matrix which represents it, we see that $\det I=1$, and we can use this to compute the determinants of any elementary operator $e_{j}$ $j=1,2,\ldots,n$. If $e_{j}$ is an elementary reflection (single row interchange), then $\det e_{j}=1$ since row (or column) interchanges only change the sign of the determinant. If $e_{j}$ is an elementary shear (single linear combination of rows), then $\det e_{j}=1$ since taking linear combinations of the rows (or columns) in a matrix does not affect the value of its determinant. Finally, if $e_{j}$ is an elementary dilation or coordinate reflection (single row scalar multiplication) by $\delta\neq0$, then $\det e_{j}=\delta_{j}$ since the determinant function is multilinear on the rows of a matrix. Using the identity $\det AB=\det A\det B$ then shows $$\det L=\prod_{j=1}^{n}\det e_{j}=\delta$$ where $\delta$ is the product of all the individual component dilations. Thus, we see that it is enough to verify the theorem for when $L$ is an elementary operator. To do this, first consider a cube $Q$ of side length $\ell$. If $L$ is an elementary reflection about a coordinate plane (e.g. the line $x_{i}=x_{k})$), then $L(Q)=Q$ since all subintervals of $Q$ are of uniform length $\ell$, and so no rotation occurs. Hence $$m(L(Q))=m(Q)=1\cdot m(Q)=\det Lm(Q).$$ In the case $L$ is an elementary dilation by $\delta=(0,...,\delta_{j},\ldots,0)$, only one of the subintervals of $Q$ is dilated, e.g. $[a_{j},b_{j})\mapsto [\delta_{j}a_{j}, \delta_{j}b_{j})$, and if $\delta_{j}<0 databloggerescapeda_="" databloggerescapedalso="" databloggerescapeddelta_="" databloggerescapedis="" databloggerescapedj="" databloggerescapedreflected="" databloggerescapedsince="" databloggerescapedthen="">\delta_{j} b_{j}$) about the orthogonal coordinate axes. But for cubes, such a reflection amounts to nothing more than a translation by $h=(0,\ldots,\pmI_{j},\ldots)$ (the sign depends on $I_{j}$ and $h=0$ if $\delta>0$) as no rotation can occur . Therefore $$m(L(Q))=m(\delta Q_{h})=m(\delta Q)=I_{1}\times\ldots\times \delta_{j}I_{j}\times\ldots\times I_{d}=\delta_{j}m(Q)=\det L m(Q).$$ The case when $L$ is an elementary shear is the most difficult, since rotations always occur, in addition to a deformation of shape. However, the deformation is predictable for a cube, and always results in a $d$ dimensional parallelogram (for example, a parallelepiped when $d=3$). Since such objects are the union of two almost disjoint $d$ dimensional triangles (along their diagonal), and the measure of a triangle agrees with the usual formula $\frac{1}{2}bh$ (where $b$ is computed from $d1$ sides) for computing the volume of a $d$simplex (we verified this in exercise 1.1.8), and each side length of the cube is the same, we have that $$ m(L(Q))=m(T_{1})+m(T_{2})=\frac{1}{2}\ell^{d}+\frac{1}{2}\ell^{d}=\ell^{d}=1\cdot m(Q)=\det Lm(Q)$$ as required. Now choose $\epsilon>0$ and let $E$ be any Jordan measurable set (note that this implies $m^{\star,(J)}(E)=m_{\star,(J)}(E)$). Then there exists collections of pairwise disjoint cubes $\{U_{i=1}\}_{i=1}^{M}$ and $\{Q_{j}\}_{j=1}^{N}$ such that both $$\bigcup\limits_{i=1}^{M}U_{i}\subset E\subset\bigcup\limits_{j=1}^{N}Q_{j}$$ and $$m(E)\epsilon\leq\sum\limits_{i=1}^{M}U_{i}\leq m(E)\leq\sum\limits_{j=1}^{N}Q_{j}\leq m(E)+\epsilon$$ hold. And, since $L$ preserves monotonicity relationships (injectivity), we obtain the estimate (where $\det L=\Delta$) \begin{align*} \Delta\Big(m(E)\epsilon\Big) &\leq \Delta\left(\sum\limits_{i=1}^{M}U_{i}\right)\\ &=\sum\limits_{i=1}^{M}m(L(U_{i})) \leq m(L(E)) \leq \sum\limits_{j=1}^{N}m(L(Q_{j}))\\ &= \Delta\left(\sum\limits_{j=1}^{N}Q_{j}\right)\\ &\leq \Delta\Big(m(E)+\epsilon\Big). \end{align*} In particular, we have the estimate $$\Bigm(L(E))\Delta m(E)\Big\leq2\Delta\epsilon,$$ and since $\epsilon$ was arbitrary, we conclude $m(L(E))=\det Lm(E)$ as required (note that boundedness of $L$ implies $\Delta<\infty$).