proof of hitting times are stopping times for right-continuous processes

Let ${(ℱ)}_{t\in 𝕋}$ be a filtration (http://planetmath.org/FiltrationOfSigmaAlgebras) on the measurable space $(\mathrm{\Omega },ℱ)$, It is assumed that $𝕋$ is a closed subset of $ℝ$ and that ${ℱ}_t$ is universally complete for each $t\in 𝕋$.

Let $X$ be a right-continuous and adapted process taking values in a metric space $E$ and $S\subseteq E$ closed.We show that

$$\tau =inf\{t\in 𝕋:X_t\in S\}$$

is a stopping time. Assuming $S$ is nonempty and defining the continuous function $d_S(x)\equiv inf\{d(x,y):y\in S\}$, then $\tau$ is the first time at which the right-continuous process $Y_t=d_S(X_t)$ hits $0$.

Let us start by supposing that $𝕋$ has a minimum element $t_0$.

If $\mathbb{P}$ is a probability measure on $(\mathrm{\Omega },ℱ)$ and ${ℱ}_t^{\mathbb{P}}$ represents the completion (http://planetmath.org/CompleteMeasure) of the $\sigma$-algebra ${ℱ}_t$ with respect to $\mathbb{P}$, then it is enough to show that $\tau$ is an $({ℱ}_t^{\mathbb{P}})$-stopping time. By the universal completeness of ${ℱ}_t$ it would then follow that

$$\{\tau \le t\}\in \bigcap _{\mathbb{P}}{ℱ}_t^{\mathbb{P}}={ℱ}_t$$

for every $t\in 𝕋$ and, therefore, that $\tau$ is a stopping time.So, by replacing ${ℱ}_t$ by ${ℱ}_t^{\mathbb{P}}$ if necessary, we may assume without loss of generality that ${ℱ}_t$ is complete with respect to the probability measure $\mathbb{P}$ for each $t$.

Let $𝒯$ consist of the set of measurable times $\sigma :\mathrm{\Omega }\to 𝕋\cup \{\mathrm{\infty }\}$ such that for every $t$ and that $\sigma \le \tau$. Then let ${\sigma }^{*}$ be the essential supremum of $𝒯$.That is, ${\sigma }^{*}$ is the smallest (up to sets of zero probability) random variable taking values in $ℝ\cup \{\pm \mathrm{\infty }\}$ such that ${\sigma }^{*}\ge \sigma$ (almost surely) for all $\sigma \in 𝒯$.

Then, by the properties of the essential supremum, there is a countable sequence ${\sigma }_n\in 𝒯$ such that ${\sigma }^{*}={sup}_n{\sigma }_n$. It follows that ${\sigma }^{*}\in 𝒯$.

For any $n=1,2,\mathrm{\dots }$ set

Clearly, ${\sigma }_1\le \tau$ and, choosing any countable dense subset $A$ of $𝕋$, the right-continuity of $Y$ gives

So, ${\sigma }_1\in 𝒯$, which implies that ${\sigma }_1\le {\sigma }^{*}$ with probability one.However, by the right-continuity of $Y$, ${\sigma }_1>{\sigma }^{*}$ whenever ${\sigma }^{*}$ is finite and $Y_{{\sigma }^{*}}>1/n$, so

This shows that $Y_{{\sigma }^{*}}=0$ and therefore ${\sigma }^{*}\ge \tau$ whenever . So, ${\sigma }^{*}=\tau$ almost surely and $\tau \in 𝒯$ giving,

So, $\tau$ is a stopping time.

Finally, suppose that $𝕋$ does not have a minimum element. Choosing a sequence $t_n\to -\mathrm{\infty }$ in $𝕋$ then the above argument shows that

$${\tau }_n=inf\{t\in 𝕋:t\ge t_n,Y_t=0\}$$

are stopping times so,

$$\{\tau \le t\}=\bigcup _n\{{\tau }_n\le t\}\in {ℱ}_t$$

as required.