Can we apply the continuous mapping theorem to indicator functions?

Question

Let $X_1,X_2,\dots$ be a sequence of i.i.d random variables with zero mean and variance 1. Define $S_n=\sum_{k=1}^n X_k$ and $B^n(t)=\frac{S_{[nt]}}{\sqrt{n}}$ for $t\in[0,1]$. Donsker's theorem tells us that the process $B^n$ converges in distribution to a standard Brownian motion $B$ in $C[0,1]$. Let $t\in[0,1]$ and $\epsilon >0$. I would like to prove that: $$\int_0^t {1}_{[0,\epsilon]}(B^n(s))ds\Rightarrow \int_0^t {1}_{[0,\epsilon]}(B(s))ds,$$ in $C[0,1]$. I have thought about using the continuous mapping theorem where we basically have to show that the maps involved are measurable and almost surely continuous. To start with, the indicator function on $[0,\epsilon]$ is almost surely continuous as its discontinuities are simply located at $0$ and $\epsilon$, which ensures the weak convergence of the indicator functions. As for the integral mapping, I believe it is continuous everywhere, which enables us to conclude. However, I am not sure whether these arguments are sufficient and/or rigorous enough to prove the result? Any ideas or comments would be greatly appreciated.

Answer 1

We can try to approximate the studied functional in the following way. For any integer $m$, define a continuous function $h_m$ on the unit interval such that $h_m=1$ on $[0,\varepsilon]$, $0$ on $[\varepsilon+1/m,1]$ and $0\leqslant h_m\leqslant 1$. Define for any positive integers $m$ and $n$ the random variables $$Y_{n,m}:=\int_0^th_m\left(B^n(s)\right)\mathrm ds\mbox{ and } Y_n:=\int_0^t\mathbf 1\left(B^n(s)\in [0,\varepsilon\right)\mathrm ds.$$ Then we have:

1. for each integer $m$, $Y_{n,m}\to \int_0^th_m\left(B(s)\right)\mathrm ds$ in distribution as $n$ goes to infinity;
2. the convergence $\int_0^th_m\left(B(s)\right)\mathrm ds\to \int_0^t\mathbf 1\left(B(s)\right)\mathrm ds$ in distribution as $m$ goes to infinity;
3. the convergence $$\lim_{m\to +\infty}\limsup_{n\to +\infty}\mathbb P\left(\left|Y_{n,m}-Y_m\right| \gt\delta\right) =0$$ for any positive $\delta$.

This is sufficient to deduce the wanted convergence (by Theorem 4.2 in Billingsley 1968). 1. follows from the continuity of the functional $G_m\colon x\in C[0,1]\mapsto \int_0^th_m\left(x(s)\right)\mathrm ds$. 2. follows from the fact that for any $u\in[0,1]$, $h_m(u)\to \mathbf 1_{[0,\varepsilon]}(u)$ . For 3., note that by Markov's inequality, \begin{align} \mathbb P\left(\left|Y_{n,m}-Y_m\right| \gt\delta\right)&\leqslant \mathbb P\left(\int_0^t\left|h_m\left(B^n(s)\right)-\mathbf 1\left(B^n(s)\in [0,\varepsilon\right)\right|\mathrm ds \gt \delta\right)\\ &\leqslant \mathbb P\left(\int_0^t\mathbf 1\left(B^n(s)\in [\varepsilon,\varepsilon+1/m]\right)\mathrm ds \gt \delta\right)\\ &\leqslant \int_0^t\mathbb P\left(B^n(s)\in [\varepsilon,\varepsilon+1/m]\right)\mathrm ds/\delta. \end{align}
We then conclude by dominated convergence.