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Suppose $\mu$ is sigma-finite measure on a space $X$, and $ f_{n}$ converge to $f$ almost everywhere. Show that there exists measurable sets $E_{n}$ where $\mu ( \cap E_{n}^{c}) = 0$ and $f_{n}$ converge uniformly to $f$ on each $ E_{n}$.

It seems we want to use egorov for this, where you have your sets of finite measure are $X_{n}$. So, $\forall n$ and for each $ X_{n}$ there is a measurable set $A_{n}$ with $ \mu (X_{n} - A_{n}) < {1}/{2^{n}} $

This way our sets can have $\mu ( \cap A_{n}^{c}) $ as small as we want but not zero. How can we make them zero?

Thank you

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You shouldn't use the same index for $f_n$ and $E_n$ and $X_n$. – Robert Israel Nov 13 '11 at 7:38
up vote 1 down vote accepted

For each $k$, by repeatedly applying Egoroff there are sets $A_{k,m} \subseteq X_k$ with $\mu(X_k \backslash A_{k,m}) \le 1/m$ on which $f_n$ converges uniformly to $f$. Let $E_m = \bigcup_{k=1}^m A_{k,m}$. Then $f_n$ converges uniformly to $f$ on $E_m$. Note that $\mu(X_k \cap \bigcap_m E_m^c) \le \mu(X_k \backslash A_{k,m}) \le 1/m$ for $m \ge k$, so $\mu(X_k \cap \bigcap_m E_m^c) = 0$, and therefore $\mu(\bigcap_m E_m^c) = 0$.

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