Let $(U_n)$ a sequence s.t. $U_n\to U$ in probability. Let $L:\mathbb R\to \mathbb R$ be a continuous function. Prove that $L(U_n)\to L(U)$. The proof goes as follow :
Step 1 : Suppose $L$ is Lipschitz on $\mathbb R$. Let $\varepsilon >0$. Then, there is $\delta >0$ s.t. $$|L(x)-L(y)|>\varepsilon \implies |x-y|>\delta \tag{E}.$$ Therefore, $$\mathbb P(|L(U_n)-L(U)|>\varepsilon )\leq \mathbb P(|U_n-U|>\delta )\underset{n\to \infty }{\longrightarrow }0.$$ Therefore $L(U_n)\to L(U)$ in probability.
Step 2 : Suppose $L$ is continuous but not uniformly continuous. Let $\varepsilon >0$, $\eta>0$ and $K>0$ s.t. $$\mathbb P(|U|\geq K)\leq \eta.$$ Since $L$ is uniformly continuous on $[-K,K]$, then $(E)$ hold whenever $|U|\leq K$.
Question : Why is this true ? Don't we need that $U\in [-K,K]$ and $U_n\in [-K,K]$ for all $n\geq N$ for some $N\in\mathbb N$ ?
Then, they conclude : therefore, $$ \mathbb P(|L(U_n)-L(U)|>\varepsilon ) $$ $$=\mathbb P(|L(U_n)-L(U)|>\varepsilon , |U|<K)+\mathbb P(|U|\geq K) $$ $$\leq \mathbb P(|U_n-U|>\delta )+\eta\underset{n\to \infty }{\longrightarrow }\eta. $$The claim follow.
So the thing I really don't understand is the question in bold. If $U_n\notin [-K,K]$ for all $n$ from a certain rang, $(E)$ doesn't hold. Any idea on why (E) is true then ?
You can use the uniform continuity on $[-K-2\varepsilon,K+2\varepsilon]$ instead. You will have $$\mathbb{P}(|L(U_{n}) - L(U)| < \varepsilon, |U| < K) \geq \mathbb{P}(|U_{n} - U| < \varepsilon, |U| < K)$$ since if $|U_{n} - U| < \varepsilon$ then $U_{n} \in [-K-\varepsilon,K+\varepsilon]$.