This is part of a proof from René Schilling's Brownian Motion. Here, $f$ is a function in $\mathscr{L}_T^2$, which is closure of the simple processes $\mathscr{S}_T$ in $L^2(\lambda_T \otimes P)$, where $\lambda_T$ is the Lebesgue measure on $[0,T]$, and simple processes are defined as a real valued stochastic process $(f(t,\cdot))_{t \in [0,T]}$ of the form $$f(t,\omega) = \sum_{j=1}^n \phi_{j-1}(\omega)1_{[s_{j-1},s_j)}(t)$$ where $n\ge 1, 0=s_0 \le s_1 \le \cdots \le s_n \le T$ and $\phi_j \in L^\infty(\mathscr{F}_{s_j})$. In the proof below, $\tau$ is a $\mathscr{F}_t$ stopping time and without loss of generality we may assume $\tau = \tau \wedge T$.
My question is: How do we use the dominated convergence here? Since $\tau \le \tau_j$ for all $j$, I don't see how we have $f(s)1_{[\tau,\tau_j)}(s) \to 0$ a.e. to use dominated convergence. I would greatly appreciate any help.
I assume you're interested in the last part, that is from $$ 4 \mathbb E \left[ \int_0^T |f(s)1_{[\tau,\tau_j)}(s)|^2ds \right] $$ go to $0$ via dominated convergence.
Note that your function under integral is bounded by $|f|^2$, which by assumption is a limit of simple processess in the norm $||\cdot ||_2$ of $L^2( \lambda_T \otimes \mathbb P)$, hence $f$ is in $L^2(\lambda_T \otimes \mathbb P)$, too.
You have integral of type $$\mathbb E \left [ \int_0^T (\cdot)d\lambda_T \right ] = \int_{\Omega \times [0,T]} (\cdot) d(\mathbb P \otimes \lambda_T) $$ and by above, $|f|^2$ is integrable on $\Omega \times [0,T]$ with respect to $\mathbb P \otimes \lambda_T$ measure. Hence you have a function that dominates your sequence.
Now, note that $\tau\le \tau_j := \frac{[2^j\tau] +1}{2^j} \le \frac{2^j\tau+1}{2^j} \to \tau$, so you have pointwise limit $\tau_j(\omega) \to \tau(\omega)$ and that means $1_{[\tau(\omega),\tau_j(\omega))}(s) \to 0$ for any $s \neq \tau(\omega)$ (so $\mathbb P \otimes \lambda_T$ almost surely, because by Fubinii $(\mathbb P \otimes \lambda_T)( \{(\omega,s) : \tau(\omega)=s\}) = \int_\Omega \int_{\{\tau(\omega)\}}dsd\mathbb P(\omega) = 0$. Hence $|f(s)1_{[\tau(\omega),\tau_j(\omega))}(s)|^2 \to 0$ almost surely with respect to $\mathbb P \otimes \lambda_T$ and hence $$ 4 \mathbb E \left[ \int_0^T |f(s)1_{[\tau,\tau_j)}(s)|^2ds \right] = 4 \int_{\Omega \times [0,T]} |f(s)1_{[\tau(\omega),\tau_j(\omega))}(s)|^2 d(\mathbb P \otimes \lambda_T)(\omega,s) \to 0$$