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If $\mu$ is $\sigma$-finite and $f_n\rightarrow f$ a.e., there exists $E_1,E_2,\ldots\subset X$ such that $\mu\left(\left(\bigcup_{1}^{\infty}E_j\right)^{c}\right)$ and $f_n\rightarrow f$ uniformly on each $E_j$.

Proof: Since $\mu$ is $\sigma$-finite, then there exists an $$X = \bigcup_{1}^{\infty}E_j \ \ \text{where} \ \ E_j\in M \ \ \text{and} \ \ \mu(E_j) < \infty \ \forall j$$ Then by continuity from below we have $E_1,E_2,\ldots\subset X$ such that $\mu\left(\bigcup_{1}^{\infty}E_j\right) = \lim_{n\rightarrow\infty}\mu(E_j)$

I am not really sure where to go from here, any suggestions is greatly appreciated.

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Since $X$ is $\sigma$-finite, let $X = \cup_{1}^{\infty}V_{n}$ with $\mu(V_{n}) < \infty$. By Egoroff's Theorem, on each $V_n$ we have a $E_n$ such that $f_n \rightarrow f$ uniformly on $E_n$ and $\mu(E_n^{c}) < 2^{-n}\epsilon$. Then we have that $\mu((\cup_{1}^{\infty}E_j)^{c}) = \mu(\cap_{1}^{\infty}E_j^{c}) < 2^{-n}\epsilon, \ \forall N$. So that $\mu((\cup_{1}^{\infty}E_j)^{c}) = 0$ and $f_n \rightarrow f$ uniformly on each $E_n$.

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    $\begingroup$ Shouldn't $\mu(E_n^c)$ should be $\mu(V_n\backslash E_n)$? That makes this slightly more complicated. $\endgroup$ Dec 9 '16 at 1:07
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Let $X = \bigcup_{n=1}^{\infty} X_n$ where $(X_n)$ is a sequence of disjoint sets with finite measures.
By Egoroff's theorem, choose $E_{n, m}\subseteq X_n$ such that $\mu(X_n\setminus E_{n,m}) < 2^{-n}2^{-m}$, $E_{n, m}\subseteq E_{n, m+1}$, and $f_n \rightarrow f$ uniformly on $E_{n,m}$.
For each $m \in \mathbb N$, $$\mu\left(\bigcap_{n=1}^{\infty} E_{n,m}^c \right) \leq \mu\left( \bigcup_{n=1}^{\infty}X_n \setminus E_{n,m} \right) \leq \sum_{n=1}^{\infty}2^{-n}2^{-m} = 2^{-m}$$ and $$\bigcap_{n=1}^{\infty} E_{n,m}^c \supseteq \bigcap_{n=1}^{\infty} E_{n,m+1}^c$$ Therefore, $$\mu\left(\left( \bigcup_{m=1}^{\infty}\bigcup_{n=1}^{\infty} E_{n,m} \right)^c\right) = \mu\left(\bigcap_{m=1}^{\infty}\bigcap_{n=1}^{\infty} E_{n,m}^c\right) = \lim_{m\rightarrow \infty}\mu\left(\bigcap_{n=1}^{\infty} E_{n,m}^c\right) = 0$$

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