Consider the following claim in a book.
Let $X$ be an optional process and $T$ a stopping time. Then $X^T$ is also an optional process.
The proof is based on the fact, that $X_T 1_{T<\infty}$ is $\mathcal{F}_T$-measurable. By a monotone class argument, it is enough to proof the fact in the case when $X$ is càdlàg and adapted. Thus, we have to proof that $X^T$ is adapted and càdlàg. The càdlàg-property is obvious. Regarding the second property, the author claims that it follows from the above lemma because $X^T_t=X_T 1_{\{T\leq t\}}+X_t 1_{\{t< T\}}$. My question is why? Clearly, $X_t 1_{\{t< T\}}$ is $\mathcal{F}_t$ measurable, but why is $X_T 1_{\{T\leq t\}}$ also $\mathcal{F}_t$ measurable?
My idea was to simply to define a stopping time $T'=T\wedge t$. Then $X^T_t=X_{T'}=X_{T'}1_{T'<\infty\}}$ is $\mathcal{F}_{T'}$ measurable. Since $T'\leq t$, we have $\mathcal{F}_{T'}\subseteq \mathcal{F}_t$. Hence $X^T_t$ is $\mathcal{F}_t$ measurable. Is this correct?
Remember that an event $A$ is in $\mathcal F_T$ if and only if $A\cap\{T\le t\}\in\mathcal F_t$ for all $t\ge 0$. This extends to random variables by simple function approximation: A r.v. $Y$ is $\mathcal F_T$-measurable if and only if $Y1_{\{T\le t\}}$ is $\mathcal F_t$-measurable for each $t\ge 0$. Now apply this to the r.v. $Y= X_T1_{\{T<\infty\}}$.
On the other hand, your simpler approach is correct as well.