If $\mu$ is $\sigma$ finite and $f_n \rightarrow f$ a.e, there exists $E_1,E_2, \ldots \subset X$ such that $\mu((\bigcup_{1}^{\infty}E_j)^{c})=0$ and $f_n \rightarrow f$ uniformly on each $E_j$

My attempt : $f_n \rightarrow f$ a.e thus $A:= \lbrace x \mid (f_n \nrightarrow f ) \rbrace $ then $\mu(A)=0$ Now let $E_1,E_2, \ldots \subset X$ such that $\bigcup_{1}^{\infty}E_j=X$ from $\sigma$-finiteness of measure , $\mu(E_j) < \infty$$ \forall $ $j$

Thus now by Egoroff theorem for every $\epsilon$ $\exists$ $E_{j_k} \subset E_j$ such that $\mu(E_{j_k})<\epsilon$ and $f_n \rightarrow f$ uniformly on each $E_{j_k}^{c}$

Now I can see I need to take union of this subsequence of $E_j$ but I dont know how to do so.

I am stuck here Please help.


1 Answer 1


For each $j,n\in \Bbb{N}$, Egoroff yields a subset $E_j^k \subset E_j$ with $\mu(E_j \setminus E_j^k) <1/k$ on which the convergence is uniform.

We now have

$$ \mu( E_j \setminus \bigcup_{\ell,n} E_\ell^n) \leq \mu (E_j \setminus E_j^n) <1/n $$

for all $n\in \Bbb{N}$, so that the left hand side is a nullset. As a countable union of nullsets, so is

$$ \Omega \setminus \bigcup_{j,n} E_j^n = (\bigcup_{j,n} E_j^n)^c. $$

Since $(E_j^n)_{j,n}$ is a countable family, we are done.


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