# Stopping time that depends on other stopping time

Consider a random variable $\tau$:

$$\tau := \inf \left\{ {k = \sigma , \ldots ,T:S_k \ge u} \right\}\bigwedge T,$$ where $\sigma$ is a stopping time, $S_k$ - stochastic process, $T, u$ are constants and $a\wedge b=\min(a,b)$. I would like to prove that $\tau$ is a stopping time, i.e. I need to show that $\{ \tau\le k\} \in {F_k}$, where ${F_k}$ is natural filtration.

I tried to present $\tau$ as $\{ \tau \le k\} = \mathop \bigcup \limits_{j = \sigma }^k \{ \tau = j\}$ and to show that $\{ \tau = j\}\in F_j, {F_j} \subset {F_k}$, so I can conclude that $\{ \tau\le k\} \in {F_k}$. But can I really use a stopping time $\sigma$ in union operator as starting index? If not, what is the correct way to prove that $\tau$ is a stopping time?

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Let $F$ denote the filtration $(F_n)_{n\geqslant0}$ and $S$ the process $(S_n)_{n\geqslant0}$. The result follows from the fact that, for every $0\leqslant k\leqslant T-1$, $$[\tau\leqslant k]=\bigcup_{i=0}^kA_i^k\qquad\text{with}\qquad A_i^k=[\sigma=i]\cup\bigcup_{j=i}^k[S_j\geqslant u].$$ Fix $0\leqslant k\leqslant T-1$. Then, for every $0\leqslant i\leqslant k$, $[\sigma=i]\in F_i$ because $\sigma$ is an $F$-stopping time, and $F_i\subset F_k$, hence $[\sigma=i]\in F_k$. For every $i\leqslant j\leqslant k$, $[S_j\geqslant u]\in F_j$ because $S$ is $F$-adapted, and $F_j\subset F_k$, hence $[S_j\geqslant u]\in F_k$.
Thus, $A_i^k\in F_k$ for every $0\leqslant i\leqslant k$, which proves that $[\tau\leqslant k]\in F_k$.
Finally, for every $k\geqslant T$, $[\tau\leqslant k]=\Omega\in F_k$, hence $\tau$ is an $F$-stopping time.