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Let $\left\{f_{n}\right\}$ be a sequence of measurable functions on the real line $\mathbb{R}$, and $f_n\rightarrow f$ almost everywhere. Prove that there exists a sequence of measurable sets $\left\{E_{k}\right\}$ such that the Lebesgue measure of $\mathbb{R}\setminus\bigcup^{\infty}_{k=1}E_{k}$ is zero, and $f_{n}\rightarrow f$ uniformly on each $E_{k}$.

The first thing occurs to me is the Egoroff's theorem. So my thought was to construct some bounded sets and select subsets of it using Egoroff's theorem to finish the proof. But I don't know how to choose these sets.

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Let $\epsilon>0$ and $\mathbb{R}=\cup_n (-n,n)=\cup_n I_n$.

Now, by Egorov's theorem, for every $k$ there exist a closed interval $E_k^n$ such that $|I_n-E_k^n|<\frac{\epsilon}{2^k}$

Note that for each $n$, $m(E_k^n)$ can be increasing and then $m(I_n-E_k^n)$ is decreasing in k

$|\mathbb{R}-\cup_{n,k}E_k^n|=|(\cup_n(\cap_k( I_n-E_k^n))|=|\cup_n(\cap_k(I_n-E_k^n))|\leq \sum_n(\lim_k |I_n-E_k^n|)=0$.

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  • $\begingroup$ How beautiful your proof is! Appreciate it. $\endgroup$ – mike Jul 24 '16 at 8:30
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    $\begingroup$ As written the solution is incorrect as the last step does not converge to zero. You should take the bound $\frac{\epsilon}{k2^n}$ instead of the bound $\frac{\epsilon}{k}$ $\endgroup$ – Ali Jul 24 '16 at 14:56

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