# Proof of convergence in distribution

I want to prove that a sequence of random variables $X_n$ converges in distribution to $N(0,1)$, if we have the following condition for an arbitrary $\epsilon>0$: $$(1-\epsilon)Y_n \le X_n \le (1+\epsilon)Z_n$$ with $Y_n,Z_n \overset d\longrightarrow N(0,1)$.

Any ideas/suggestions?

• Do you mean for all $\epsilon>0$ or just for a particular one? – Alex R. Sep 15 '14 at 16:56
• $\epsilon>0$ is arbitrary, therefore proving it for a particular one would be sufficient. – Scooby Sep 15 '14 at 17:02

I will rather write $Y_{n,\varepsilon}$ and $Z_{n,\varepsilon}$ in order to make appear the crucial dependence on $\varepsilon$. The letter $N$ denotes a random variable whose distribution is standard normal.
Fix $t\in\mathbb R$. Then for each $n$, $\varepsilon$, $$\mu\{X_n\leqslant t\}\leqslant \mu\{(1-\varepsilon)Y_{n,\varepsilon}\leqslant t\},$$ hence for each fixed $\varepsilon$. $$\limsup_{n\to \infty}\mu\{X_n\leqslant t\}\leqslant \mu\{(1-\varepsilon)N\leqslant t\}.$$ Since $(1-\varepsilon_k)Y\to Y$ almost surely when $\varepsilon_k\to 0$, we have, $$\limsup_{n\to \infty}\mu\{X_n\leqslant t\}\leqslant \mu\{N\leqslant t\}.$$ Similarly, using the other inequality, we get $$\liminf_{n\to \infty}\mu\{X_n\leqslant t\}\geqslant \mu\{N\leqslant t\}.$$