Understanding a proof of sum of stopping times

Question

Lemma:Let $\sigma$ and $\tau$ be stopping times. If $\sigma,\tau\geqslant 0$ then $\sigma+\tau$ is also a stopping time.

Proof:

Let $t\in I$. By a previous result, $\tau\wedge t$ and $\sigma\wedge t$ are stopping times for any $t\in I$(where $\tau\wedge t$ reads $\min\{\tau, t\}$). In particular $\{\tau\wedge t< s\}\in\mathscr{F}_s\subset \mathscr{F}_t$ for any $s\leqslant t$. On the other hand, we have $\tau\wedge t\leqslant s$ for $s>t$. Hence $\tau'=(\tau\wedge t)+\mathbb{1}_{\{\tau>t\}}$ and $\sigma'=(\sigma\wedge t)+\mathbb{1}_{\{\sigma>t\}}$ and thus $\tau' +\sigma'$ are $\mathscr{F}_t$ measurable. We conclude $\{\tau+\sigma\leqslant t\}=\{\tau'+\sigma'\leqslant t\}\in\mathscr{F}_t$.

Observation:

I have been thinking about this proof. But I cannot make sense of $\tau'=(\tau\wedge t)+\mathbb{1}_{\{\tau>t\}}$ and $\sigma'=(\sigma\wedge t)+\mathbb{1}_{\{\sigma>t\}}$ once both expressions in case $\tau>t$ are going to equal $t+1>t$. So how come $\tau'$, $\sigma'$ and its sum be $\mathscr{F}_t$ measurable?

Question:

Can someone explain me this proof? Why is $\tau'+\sigma'$ measurable?

Thanks in advance!

Answer 1

Step 1: $\tau' := (\tau \wedge t)+ 1_{\{\tau>t\}}$ is $\mathcal{F}_t$-measurable.

We have

$$\{\tau \wedge t \leq s\} \in \mathcal{F}_s \subseteq \mathcal{F}_t, \qquad s \leq t,$$

and

$$\{\tau \wedge t \leq s\} = \Omega \in \mathcal{F}_t, \qquad s>t$$

which means that $\tau \wedge t$ is $\mathcal{F}_t$-measurable. The random variable $1_{\{\tau>t\}}$ is $\mathcal{F}_t$-measurable as $\{\tau>t\} =\{\tau \leq t\}^c \in \mathcal{F}_t$. Hence, $\tau'$ is $\mathcal{F}_t$-measurable as sum of $\mathcal{F}_t$-measurable random variables.

In exactly the same manner we find that $\sigma' := (\sigma \wedge t) + 1_{\{\sigma>t\}}$ is $\mathcal{F}_t$-measurable.

Step 2: $\tau(\omega) \leq t \iff \tau'(\omega) =\tau(\omega) \iff \tau'(\omega) \leq t$

Indeed: If $\tau(\omega) \leq t$, then $\tau'(\omega) = \tau(\omega) + 1_{\{\tau>t\}}(\omega)=\tau(\omega)$. Hence, $$\{\tau \leq t\} \subseteq \{\tau' =\tau\}.\tag{1}$$ On the other hand, if $\tau(\omega)>t$ then $\tau'(\omega)=t+1 >\tau(\omega)$ which means that $$\{\tau>t\} \subseteq \{\tau'\neq \tau\},$$ i.e. $$\{\tau \leq t\} \supseteq \{\tau'=\tau\}.\tag{2}$$ Combining $(1)$ and $(2)$ gives the first assertion. The proof of the second one is very similar.

In the same fashion we get $\sigma(\omega) \leq t \iff \sigma'(\omega) =\sigma(\omega) \iff \sigma'(\omega) \leq t$.

Step 3: $\{\tau+\sigma \leq t\} = \{\tau'+\sigma' \leq t\}$

Indeed: \begin{align*} \tau(\omega)+\sigma(\omega) \leq t &\iff \tau(\omega)+\sigma(\omega) \leq t, \tau(\omega) \leq t, \sigma(\omega) \leq t \\ &\stackrel{\text{Step 2}}{\iff} \tau(\omega)+\sigma(\omega) \leq t, \tau'(\omega) = \tau(\omega), \sigma'(\omega) = \sigma(\omega) \\ &\stackrel{\text{Step 2}}{\iff} \tau'(\omega)+\sigma'(\omega) \leq t, \tau'(\omega) \leq t, \sigma'(\omega) \leq t, \\ &\iff \tau'(\omega) + \sigma'(\omega) \leq t. \end{align*}

Conclusion

Since $\tau'+\sigma'$ is $\mathcal{F}_t$-measurable, it follows that

$$\{\tau+\sigma \leq t\} = \{\tau'+\sigma' \leq t\} \in \mathcal{F}_t.$$

Hence, $\tau+\sigma$ is a stopping time.