the limit of the infimum of a sequence of bounded functions that converge uniformly is equal to the infimum of the limiting function

Question

prove $ \lim_{n\to\infty}inf [f_n(x)|x\in E]=inf[f(x)|x\in E]$ where $f_n$ are bounded functions of a set $E\subset R$ that converge uniformly to a function $f$. I've looked at similar proofs on the website before, but they all require that the functions be continuous. I've managed to prove that $ \lim_{n\to\infty}inf(f_n(x)|x\in E)\leq inf(f(x)|x\in E)$, but cannot prove the other direction.

Answer 1

Suppose $|f_n(y)-f(y)| <\epsilon $ for all $n \geq m $ for all $y \in E$. For any $x \in E$ we have $f(x) \leq f_n(x)+\epsilon$ if $n \geq m$. Take infimum over $x \in E$ on both sides. We get $\inf \{f(x): x \in E\} \leq \inf \{f_n(x): x \in E\}+\epsilon$. But $\epsilon$ is arbitrary so we get $\inf \{f(x): x \in E\} \leq \inf \{f_n(x): x \in E\}$ for all $n \geq m$.