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Let's take a stochastic process $(X_t)_{0\leq t \leq 1}$ and assume that the sample paths are almost surely continuous. Let us define $S \equiv \sup_{t \in [0,1]} X_t$. How can we show that $S$ is measurable?

For example, the if we take the Brownian motion $B_t$ as our stochastic process, then given the continuity of the sample paths, we can focus on the supremum over $t \in [0,1] \cap \mathbb{Q}$, which is a countable, dense subset of $[0,1]$, and we have continuity of $B_t$, therefore the supremum is measurable (see the answer here: Measurability of the supremum of a Brownian motion).

How does almost sure continuity instead of continuity change the way of proving measurability?

I would be very grateful for any hint!

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  • $\begingroup$ You can always throw away a set of measure zero, without affecting the measure of anything else. That is, if $A$ is measurable and $B$ has measure zero then $A \cap B^c $ has the same measure as $A$. $\endgroup$
    – Michael
    Jul 4, 2016 at 2:50
  • $\begingroup$ @Michael: thank you for your comment! So if I understand your argument, then we could say the following. 1. Let $\tilde{X}_t$ be equal to $X_t$ at its continuity points. 2. Then $\tilde{S} \equiv \sup_{t \in [0,1]}\tilde{X}_t$ is measurable. 3. Given that $X_t$ and $\tilde{X}_t$ differ only on a set of measure zero, their suprema must be equal, $S=\tilde{S}$. Is there any way to make the last part slightly more rigorous? $\endgroup$
    – acuriousmind
    Jul 4, 2016 at 10:23
  • $\begingroup$ When your question said "sample paths are almost surely continuous" I interpreted that to mean: "For almost all outcomes $\omega$ in the sample space, $X_t(\omega)$ is a continuous function over $t \in [a,b]$." That is different from saying "the functions are continuous over $[a,b]$ except possibly on a set of measure zero." In other words, I interpret "almost surely" in a probability sense according to the probability measure of the stochastic process, rather than in terms of measure on the interval $[0,1]$. $\endgroup$
    – Michael
    Jul 4, 2016 at 14:27
  • $\begingroup$ @Michael: indeed, I meant the former and I think both your comment and my reply to your comment deal with this case. I would be happy to reformulate my question if you could tell me which part was ambiguous. $\endgroup$
    – acuriousmind
    Jul 4, 2016 at 21:45
  • $\begingroup$ When your response to my comment talked of "continuity points" of $X_t$, it sounded like you meant those points in the interval $t \in [0,1]$ at which $X_t$ is continuous. That made me wonder if you really meant that the functions $X_t$ were continuous at all points $t$ in the interval $[0,1]$, except possibly a subset of points in $[0,1]$ that have measure 0. $\endgroup$
    – Michael
    Jul 4, 2016 at 22:14

1 Answer 1

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We have to assume that the underlying probability space is complete; otherwise the assertion might fail.

So, suppose that $(\Omega,\mathcal{A},\mathbb{P})$ is a complete probability space and $(X_t)_{t \in [0,1]}$ a process with almost surely continuous sample paths, i.e. there exists a null set $N \in \mathcal{A}$ such that $$[0,1] \ni t \mapsto X_t(\omega)$$ is continuous for all $\omega \in \tilde{\Omega} := \Omega \backslash N$. Now

$$\tilde{X}_t(\omega) := \begin{cases} X_t(\omega), & \omega \in \tilde{\Omega}, \\ 0, & \omega \in N \end{cases}$$

defines a stochastic process on $\Omega$ with continuous sample paths, and therefore

$$\sup_{t \in [0,1]} \tilde{X}_t = \sup_{t \in [0,1] \cap \mathbb{Q}} \tilde{X}_t$$

is measurable as countable supremum of measurable random variables. On the other hand, we have

$$\tilde{S}_t(\omega) = \sup_{t \in [0,1]} \tilde{X}_t(\omega) = \sup_{t \in [0,1]} X_t(\omega)= S_t(\omega) \quad \text{for all $\omega \in \tilde{\Omega} = \Omega \backslash N$}$$

and so

$$\{S_t \in B\} = \left( \{\tilde{S}_t \in B \} \cap N^c \right) \cup \bigg( \{S_t \in B \} \cap N \bigg)$$

for any Borel set $B$. Since $N \in \mathcal{A}$ and $\tilde{S}_t$ is measurable, we know that

$$\left( \{\tilde{S}_t \in B \} \cap N^c \right) \in \mathcal{A}.$$

Moreover,

$$\left\{ S_t \in B \right\} \cap N \subseteq N$$

and since the probability space is complete, this implies

$$\left\{ S_t \in B \right\} \cap N \in \mathcal{A}.$$

Combining both considerations proves $\{S_t \in B\} \in \mathcal{A}$, and this proves the measurability of $S_t$.

Remark More generally, the following statement holds true in complete probability spaces:

Let $(\Omega,\mathcal{A},\mathbb{P})$ and $(E,\mathcal{B},\mathbb{Q})$ be two measure spaces and assume that $(\Omega,\mathcal{A},\mathbb{P})$ is complete. Let $X, Y: \Omega \to E$ be two mappings. If $X$ is measurable and $X=Y$ almost surely, then $Y$ is measurable.

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  • $\begingroup$ I don't understand your remark: what is the meaning of "random variable" without a measurability assumption? And what is the relationship between $X$ and $Y$? +1 for the rest, though. $\endgroup$
    – Ian
    Jul 4, 2016 at 17:54
  • $\begingroup$ @Ian Ah, sorry... should be clearer now. (Thanks for your upvote :)) $\endgroup$
    – saz
    Jul 4, 2016 at 17:58
  • $\begingroup$ @saz: great answer, I will need to digest it. Much appreciated! $\endgroup$
    – acuriousmind
    Jul 4, 2016 at 21:56
  • $\begingroup$ @math_student5 You are welcome. $\endgroup$
    – saz
    Jul 5, 2016 at 5:18
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    $\begingroup$ @quallenjäger Consider $\Omega := \mathbb{R}$ endowed with the dirac measure $\mathbb{P} := \delta_0$ and the $\sigma$-algebra $$\mathcal{A} := \{A \subseteq \Omega; \sharp A < \infty \, \, \text{or} \, \, \sharp (A^c)<\infty\}.$$ The process $$X_t(\omega) := \omega 1_{\{|\omega|\}}(t)$$ has $\mathbb{P}$-almost surely continuous sample paths (the exceptional null set is $N:= \mathbb{R} \backslash \{0\}$). However, $$\{S_1 \in (0,1)\} = (0,1)$$ and therefore clearly $$\{S_1 \in (0,1)\} \cap N= (0,1) \notin \mathcal{A}$$ $\endgroup$
    – saz
    May 19, 2017 at 5:07

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