A user from the Quant StackExchange recently asked why the regularity of condition of Brownian motion, namely almost sure continuity, is what it is: almost sure? Why can't this be upgraded to Brownian motion being surely continuous?

The answer to the latter question is that, actually, it can and very often is. The answer to the former question is that stipulating almost sure continuity is required in order to make the defining conditions of Brownian motion axiomatic, while still encompassing all of the methods of construction.

Indeed, the most common construction of Brownian motion (or at least the most direct) is through an application of Kolmogorov's extension theorem (the details of this approach can be found in Durrett). But due to technical issues arising from measure theory (which are actually quite natural), the resulting construction leads to realizations of Brownian motions that are discontinuous.

On the other hand, the approach to constructing Brownian motion from the limit of scaled random walks actually leads to surely continuous realizations. There are two available routes one can go when having this approach in mind: (a) construct Brownian motion paths directly (i.e. pointwise) from scaled random walks (one common way to do this is by appropriately specifying Brownian motion on the dyadic intervals, interpolating between, and taking limits) or (b) construct Brownian motion by obtaining the Wiener measure on the space of continuous functions beginning at the origin from the induced measures on this space obtained from the scaled random walks on $\mathbb{Z}^{\infty}_{2},$ the space of sequences with values of $-1$ or $1$.

The user also asked whether an explicit example of a discontinuous Brownian motion path could be exhibited. The following is my complete answer to this and the above questions.

____________________________

Exhibiting a counter-example is straight-forward enough. For example, let $B_{t}(\omega)$ be a Brownian motion and $\mathcal{T}(\omega)$ a stopping time on $(\Omega,\mathbb{P})$ with a continuous distribution.

Then with

$$B'_{t}(\omega)=\left\{\begin{array}{ll}B_{t}(\omega),&t\neq\mathcal{T}(\omega)\\B_{t}(\omega)+1,&t=\mathcal{T}(\omega),\end{array}\right.$$

$B'_{t}(\omega)$ satisfies (1) and (2) below, but is discontinuous precisely when $t=T(\omega)$. Therefore, $B_{t}(\omega)$ is a particular realization of Brownian motion that is not everywhere continuous.

There are lots of other ways to obtain a "bad" Brownian motion. Another example is

$$B'_{t}(\omega)=B_{t}(\omega)\mathbb{1}_{\{B_{t}(\omega)\;\text{irrational}\}},$$

but this is less straight-forward to prove.

____________________________

The reason for stipulating almost sure continuity has to do with the way one constructs Brownian motion, and the issue can be completely dispensed with dependent on one's approach.

The usual presentation in finance texts is the abstract one, namely given a probably space $(\Omega,\mathbb{P})$, one has a Brownian motion $B_{t}(\omega)$ on this space if

- For every set of times $0\leq t_{1}<t_{2}<\ldots<t_{n}$ the increments $B_{t_{1}},B_{t_{2}}-B_{t_{1}},\ldots,B_{t_{k}}-B_{t_{k-1}}$ form a mutually independent set of random variables on $(\Omega,\mathbb{P}).$
- The increments above are normally distributed with mean $0$ and variance $\Delta t$.
- For almost every $\omega\in\Omega$ the path $t\mapsto B_{t}(\omega)$ is continuous.

Indeed, if we start with $(\Omega,\mathbb{P})$ satisfying the above and let $\mathcal{P}$ denote the collection of continuous functions $[0,\infty)\to\mathbb{R}$ with $p(0)=0$, then we get from (3) the inclusion map

$$\mathcal{i}:\Omega\to\mathcal{P},$$

defined on a set $\Omega'\subset\Omega$ of full measure, and the push-forward measure of $\mathbb{P}$ onto $\mathcal{P}$ under this inclusion map turns out to be equal to the Wiener measure $\mathbb{W}$ on $\mathcal{P}$, which is unique.

Conversely, one can construct $(\mathcal{P},\mathbb{W})$ directly by starting with the set $\mathcal{P}$ (where every element of this set is continuous a priori) and demonstrating that the measures $\mu_{N}$ on $\mathbb{Z}^{\infty}_{2}$ arising from the appropriately scaled random walks $S_{t}^{N}(\omega)$ ($\omega\in\mathbb{Z}^{\infty}_{2})$ induce a collection of tight measures on $\mathcal{P}$ which converge weakly to $\mathbb{W}$:

$$\mu_{N}\Longrightarrow\mathbb{W}\;\text{(weakly)}$$

One then defines

$$\tilde{B}_{t}(\omega):=p(t)\in\mathcal{P}$$

and readily shows that under $\mathbb{W}$, $\tilde{B}_{t}$ satisfies (1)-(3) and that therefore

$$\tilde{B}_{t}(\omega)=B_{t}(\omega),$$

but that now *every* Brownian motion is continuous.

The equivalence of the implications above show the existence of Brownian motion is essentially tantamount to the existence of a Wiener measure on $\mathbb{W}$ arising from the sequence of measures arising naturally from the scaled random walks. If one starts from the goal of obtaining this measure, one gets continuity for *every* Brownian motion $p(t)=B_{t}(\omega)$.

____________________________

Other constructions of Brownian motion require us stipulate almost sure continuity due to technicalities arising from measure theory on product spaces. The quickest construction of Brownian motion in this direction is by applying Kolmogorov's extension theorem on a suitable class of processes; details can be found in Durrett.