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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.

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  • $\begingroup$ 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 $1/n$. $\endgroup$ – Apocalypse Nov 17 '18 at 7:08
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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 $\|f\|_{L^{\infty}(\mu)} \leq C$.

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