Proving if $X_n$ are uniformly integrable and $X_n \Rightarrow X$, then $EX_n \to EX$ Below is the proof from Billingsley's Convergence of Probability Measures. However, in the proof, I don't understand the final step. 
That is, we want to show that$$
\int_0^\alpha P[t<X_n<\alpha] \,\mathrm{d}t \to \int_0^\alpha P[t<X<\alpha] \,\mathrm{d}t
$$
using the bounded convergence theorem. However, to use the bounded convergence theorem, we need that $P[t<X_n<\alpha]$ converges pointwise to $P[t<X<\alpha]$. But the function used here, $\mathbb{1}_{(t,\alpha)}(x)$ is not a continuous function. So how do we show pointwise convergence here? And what does $P[X=\alpha]=0$ have to do here?


$\def\dto{\xrightarrow{\mathrm{d}}}\def\d{\mathrm{d}}$Suppose $D = \{x > 0 \mid F_X \text{ is not continuous at } x\}$, then $D$ is a countable set, and for any fixed $α > 0$ and $t \in [0, α]$,$$
P(X = t) > 0 \Longrightarrow t \in [0, α] \cap D.
$$
If $P(X = α) = 0$, then for any $t \in [0, α] \setminus D$, there is $P(X \in \partial([t, α])) = 0$. Thus $X_n \dto X$ implies$$
\lim_{n \to \infty} P(t < X_n <α) = P(t < X < α), \quad \forall t \in [0, α] \setminus D
$$
so for $t \in [0, α]$,$$
\lim_{n \to \infty} P(t < X_n <α) = P(t < X < α). \quad \mathrm{a.e.}
$$
Since for any $n \geqslant 1$,$$
0 \leqslant P(t < X_n <α) \leqslant 1, \quad \forall t \in [0, α]
$$
then the bounded convergence theorem implies that$$
\lim_{n \to \infty} \int_0^α P(t < X_n < α) \,\d t = \int_0^α P(t < X < α) \,\d t.
$$
