Proving if $X_n$ are uniformly integrable and $X_n \Rightarrow X$, then $EX_n \to EX$

Question

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?

Answer 1

We know that

\begin{equation}\tag{1}\text{$X_n \Rightarrow X$ if and only if $P(X_n \in A) \to P(X \in A)$ for all $A$ such that $P(X \in \partial A)=0$.}\end{equation}

Recall that there are at most countably many $t$ such that $P(X=t)>0$. With $\alpha$ chosen so that $P(X=\alpha)=0$ it follows that

\begin{equation}\tag{2}\text{$P(X \in \partial(t,\alpha)) = 0$ for all but countably many $t$.}\end{equation}

But a countable collection of real numbers $t$ has Lebesgue measure $0$, so by (1), (2), and the assumption that $X_n \Rightarrow X$, we have

\begin{equation}\text{$P(t < X_n < \alpha) \to P(t<X<\alpha)$ for Lebesgue-almost every $t \in [0, \alpha].$}\end{equation}

By the bounded convergence theorem we conclude that $\int_0^\alpha P(t < X_n < \alpha)dt \to \int_0^\alpha P(t<X<\alpha)dt.$

Answer 2

$\def\dto{\xrightarrow{\mathrm{d}}}\def\d{\mathrm{d}}$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$$ \lim_{n \to \infty} \int_0^α P(t < X_n < α) \,\d t = \int_0^α P(t < X < α) \,\d t. $$