# Measurability of Last Exit Time of a Discrete Time Stochastic Process

Suppose we have a discrete-time stochastic process $$\{X_t\}$$ defined on a space $$(\Omega,\mathcal{F})$$ equipped with the probability measure $$\mathbb{P}$$. Suppose we know that $$X_t \rightarrow 0$$ almost surely. Let $$T(\epsilon)$$ be defined as $$T(\epsilon)=sup\{t\in\mathbb{N}: |X_t| \geq \epsilon\}$$. My question is whether $$T(\epsilon)$$ is measurable w.r.t. the $$\sigma$$-algebra $$\mathcal{F}$$. This appears to be true, but a proof or a reference to a proof would be very helpful.

• $T(\epsilon)$ is an $\mathbb{N}$-valued random variable, not a subset of $\Omega$. What do you mean by "$T(\epsilon)$ is measurable w.r.t. the $\sigma$-algebra $\mathcal{F}$"?
– kccu
Commented Apr 8, 2019 at 22:28
• Commented Apr 8, 2019 at 22:30
• Can you clarify whether each $X_t$ is defined on $(\Omega,\mathcal{F})$ or whether the entire sequence $(X_0,X_1,X_2,\dots)$ is?
– kccu
Commented Apr 8, 2019 at 22:33
• Ah, the entire sequence is defined on $\mathcal{F}$. Commented Apr 8, 2019 at 22:37
• @kccu : Basically, given that the entire discrete stochastic process $\mathcal{X}_t$ is measurable w.r.t. the sigma algebra $\mathcal{F}$, I am interested in knowing whether the quantity $T(\epsilon)$ qualifies as a $\mathcal{F}$-measurable random variable. Commented Apr 8, 2019 at 22:41

To verify a positive integer valued random variable $$N$$ is measurable, it suffices to show the sets $$\{N=n\}$$ are measurable, for $$n\in \mathbb N$$.

$$\{T(\epsilon)=n\}=\{|X_n|\ge \epsilon\}\cap \bigcap_{t=n+1}^{\infty} \{|X_t|< \epsilon\}$$ The RHS is measurable because it is a countable intersection of measurable events.

• Thanks, that makes sense. Commented Apr 8, 2019 at 22:57
• I have an unrelated (perhaps trivial) question. Suppose $\mathbb{X},\mathbb{Y}$ are random variables on a common space, and $a,\epsilon$ are positive constants with $\epsilon < a$. Is the following true: $\mathbb{P}(\mathbb{X} \leq a-\epsilon) \leq \mathbb{P}(\mathbb{X} \leq a-\mathbb{Y} | |\mathbb{Y}| \leq \epsilon)$? This seems to be trivially true because of the obvious implication of events, but am I overlooking any subtlety induced by the conditional operator? Commented Apr 8, 2019 at 23:25
• You're welcome. If an answer is helpful, you should mark it as accepted by clicking the check mark. Also, your second comment would be better asked as a separate question. Commented Apr 9, 2019 at 0:03
• Okay, let me ask it as a separate question then: any comments on the same would be appreciated, thanks. Commented Apr 9, 2019 at 0:30