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.


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 $E_n$.

  • 3
    $\begingroup$ Shouldn't $\mu(E_n^c)$ should be $\mu(V_n\backslash E_n)$? That makes this slightly more complicated. $\endgroup$ – user3281410 Dec 9 '16 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.