# Showing that the almost uniform limit of functions with bounded $L_\infty$ norms is in $L_\infty$

Suppose that $$(f_n)$$ is a sequence of functions for which there exists a finite constant $$C$$ such that the $$L_\infty$$ norm of $$f_n$$ is less than or equal to $$C$$ for all $$n$$. Suppose further that $$f_n$$ converges "almost uniformly" to a function $$f$$, i.e. for every $$\epsilon>0$$ there exists a set $$E$$ whose complement has measure less than $$\epsilon$$ such that $$f_n$$ converges uniformly to $$f$$ on $$E$$. Then show that $$f$$ is in $$L_\infty$$.

I'm not sure how to approach this. One thought I had is using the result that $$L_\infty$$ convergence is equivalent to uniform convergences almost everywhere, i.e. there exists a set $$E$$ whose complement has measure $$0$$ such that $$f_n$$ converges to $$f$$ uniformly on $$E$$. But I'm not sure how to use the boundedness of the $$L_\infty$$ norms and almost uniform convergence to prove uniform convergence almost everywhere.

• Perhaps you can try to use the diagonal method to find a subsequence $f_{n,1/n}$ where $1/n$ means the $\epsilon$ in your problem is set to be $1/n$. – Apocalypse Nov 17 '18 at 7:08

By assumption there exists for any $$k \in \mathbb{N}$$ a measurable set $$E_k$$ such that $$\mu(E_k^c) \leq \frac{1}{k}$$ and $$f_n \to f$$ uniformly on $$E_k$$. For $$x \in E_k$$ we clearly have

$$|f(x)| = \lim_{n \to \infty} |f_n(x)| \leq C.$$

Since $$k \in \mathbb{N}$$ is arbitrary, this shows that

$$|f(x)| \leq C \quad \text{for any} \, \, x \in X := \bigcup_{k \in \mathbb{N}} E_k.$$

As

$$\mu \left( X^c \right) \leq \lim_{k \to \infty} \mu(E_k^c)=0,$$

we conclude that $$\|f\|_{L^{\infty}(\mu)} \leq C$$.