$f_n \to f$ almost uniform if and only if $\lim_{n \to \infty} \mu ( \cup_{m \geq n} |f_m(x) - f_n(x)| \geq \epsilon ) = 0 $

Question

Show that $f_n \to f$ almost uniform if and only if for all $\epsilon >0$ $\lim_{n \to \infty} \mu ( \cup_{m \geq n} |f_m(x) - f_n(x)| \geq \epsilon ) = 0 $

I have showed the first direction, but I'm having problems with $\Longleftarrow$

The definition of $f_n$ converges almost uniformly to $f$, is that for every $\epsilon > 0$, there exists a set $E_\epsilon$ with $\mu(E_\epsilon) < \epsilon$ such that $f_n$ converges uniformly to $f$ on the complement of $E_\epsilon$. That is, for all $x \in E_\epsilon^c$, we have $\lim_{n \to \infty} |f_n(x) - f(x)| = 0$.

Can someone help me?

Answer 1

Hints: Replacing $\epsilon$ by $\epsilon /k$ in the hypothesis we see that there exists $n_k$ such that $\mu (\cup_{m \geq n} |f_m(x) - f_n(x)| \geq \epsilon/k ))<\epsilon/2^{k}$ whenever $m \geq n \geq n_k$.

Let $E_k= ((\cup_{m \geq n_k} |f_m(x) - f_n(x)| \geq \epsilon/k ))$. Then $\mu (\cup_kE_k)\leq \epsilon$. For $x$ in the complement of $\cup_k E_k$, the sequence $(f_n(x))$ is Cauchy. Let $f(x)=\lim f_n(x)$ for such $x$ (and $0$ for all other $x$). Check that $f_n \to f$ uniformly on the complement of $\cup_k E_k$.