In the course of proving Wald's second identity $E(B^2_T)=E(T)$, where $(B_t)_{t\geq0}$ is the Brownian motion and $T$ is a stopping time with E(T)<\infty, I got stuck with the following problem. The notation used is $T \wedge n = \min(T,n)$.
I already have $E(T)=E(\lim_{n \to \infty} T \wedge n)\\ =\lim_{n \to \infty} E(T \wedge n)\\ =\lim_{n \to \infty} E(B^2_{T\wedge n}).$ by monotone convergence and the optional stopping theorem.
Furthermore, by the Lemma of Fatou E(B^2_T)=E(\lim_{n \to \infty} B^2_{T\wedge n})\\ \leq \liminf_{n \to \infty} E(B^2_{T\wedge n})\\ = E(T) < \infty.
And now I am stuck with the other direction. I tried to use the dominant convergence theorem to exchange the limits in $E(\lim_{n \to \infty} B^2_{T\wedge n})=\lim_{n \to \infty}E(B^2_{T\wedge n})$, but I can't find a suitable integrable dominating function for $B^2_{T\wedge n}$.
Doob's inequality for stopping times yields E(\sup_{t \geq 0} B^2_{t \wedge T\wedge n})\leq 4 E(B^2_{T\wedge n}) \leq 4 E(T) <\infty, but what I need is E(\sup_{n \in \mathrm{N}} B^2_{T\wedge n})<\infty.