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

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?

• @geetha290krm But how can I include the $f(x)$ term? Commented Sep 11, 2023 at 5:42

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 $$\cup_k E_k$$.