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From Wikipedia

Let $(Ω, Σ, P)$ be a probability space, let $T$ be some interval of time, and let $X : T × Ω → S$ be a stochastic process. For simplicity, the rest of this article will take the state space $S$ to be the real line $\mathbb{R}$, but the definitions go through mutatis mutandis if $S$ is $\mathbb{R}^n$, a normed vector space, or even a general metric space.

Given a time $t ∈ T$, $X$ is said to be continuous with probability one at $t$ if $$ \mathbf{P} \left( \left\{ \omega \in \Omega \left| \lim_{s \to t} \big| X_{s} (\omega) - X_{t} (\omega) \big| = 0 \right. \right\} \right) = 1. $$

$X$ is said to be sample continuous if $X_t(ω)$ is continuous in $t$ for $P$-almost all $ω ∈ Ω$.

If I am correct, $T$ is an interval of $\mathbb{R}$.

Obviously if $X$ is sample continuous, then $X$ is continuous at every $t \in T$.

I was wondering if the reverse is true? I guess no, because uncountable union of null measurable subsets (each for each $t \in T$) may not be measurable, and even if it is measurable, its measure may not be zero?

Thanks and regards!

Tim
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  • @StefanHansen: sample continuity is actually a.s. sample continuity. (see the definition). Why would you think of them being same? – Tim Feb 07 '13 at 08:54
  • @StefanHansen:But for different $t$, this set may be different. Shall we take their intersection across all $t$? Will the interesection still be measurable, and has measure 1 or less than 1? – Tim Feb 07 '13 at 13:33
  • Sorry I misread it - nevermind my comments. – Stefan Hansen Feb 07 '13 at 13:36

1 Answers1

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No. For a counterexample, let $(Ω, Σ, P)$ be $[0,1]$ with Lebesgue measurable sets and Lebesgue measure, $T:={\Bbb R}$, and $$ X_t(\omega):=\left\{\begin{array}{ll}0, & t<\omega,\\ 1, & t\ge\omega.\end{array}\right. $$ Then the process is a.s. continuous at any given time but has no continuous sample paths as $t\mapsto X_t(\omega)$ changes from $0$ to $1$ at $t=\omega$.

A less artificial example is the telegraph process, in which $X_t$ is always either $0$ or $1$ but flips from $0$ to $1$ and from $1$ to $0$ intermittently.

David Moews
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