$f_n \rightarrow_{\mu} f$ in measure , $g$ uniformly continuous $\Rightarrow$ $g\circ f_n \rightarrow_{\mu} g \circ f$

Question

Let $f_n$ converge to $f$ in measure: $f_n \rightarrow_{\mu} f$.

Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be uniformly continuous.

I want to show that $g\circ f_n \rightarrow_{\mu} g \circ f$.

I use the definitions:

$f_n \rightarrow_{\mu} f \Leftrightarrow \lim_{n \rightarrow \infty}\mu (\{x \in X |f_n(x)-f(x) |>\epsilon\})=0$.

$g$ is uniformly continuous $\Leftrightarrow \forall \epsilon>0 \exists \delta>0 \forall x,x_0 \in D: |x-x_0|<\delta \Rightarrow |g(x)-g(x_0)|<\epsilon$

So I want to show that $\lim_{n \rightarrow \infty}\mu (\{x \in X |(g \circ f_n)(x)-(g\circ f)(x) |>\epsilon\})=0$.

Answer 1

Fix $\epsilon$ That is $\delta$ s.t. $|x-y|<\delta \Rightarrow |g(x)-g(y)|\leq\epsilon$.

In further, $ A_n:=\{x| |g(f_n(x)) -g(f(x)) | > \epsilon \}$ so that $x\in A_n \Rightarrow |f_n(x)-f(x)|\geq \delta $

Hence $$ \mu\ A_n \leq \mu\ \{x| |f_n(x)-f(x)|\geq \delta \} \rightarrow 0 $$