Let $(E,\Sigma, \mu)$ be a $\sigma$-finite measurable space (i.e., $E=\bigcup_{k \in \mathbb N} A_k$ where $\mu(A_k) < \infty$ for each $k$). Let $(f_n)_{n \geq 1},f:E \to \overline{R}$ be a sequence of measurable, a.e. finite functions such that $f_n \to f$ a.e. on $E$. Show that there exists a sequence $(E_i)_{i \geq 1}$ of measurable sets on $E$ such that $$\mu(E \setminus \bigcup_{i \geq 1} E_i)=0, f_n \rightrightarrows f \space \text{on} \space E_i \space \text{for each i}$$

I've tried to show this result but I got stuck at one part, I'll write what I could do:

In each $A_k$ I can apply Egorov's theorem, so for each $j \in \mathbb N$, there exists $B_{k,j}$ measurable subset of $A_k$ with $\mu(A_k \setminus B_{k,j})<\dfrac{1}{j}$ and $f_n \rightrightarrows f$ on $B_{k,j}$. Then $(B_{k,j})_{j}$ is a sequence of measurable subsets of $A_k$ with $$\mu(A_k \setminus \bigcup_{j \geq 1} B_{k,j})<\mu(A_k\setminus B_{k,j})<\dfrac{1}{j} \space \text{for each} j \in \mathbb N$$

But then $\mu(A_k \setminus \bigcup_{j \geq 1} B_{k,j})=0$ and $f_n$ converges uniformly to $f$ on each $B_{k,j}$.

I defined $E_k=\bigcup_{j \geq 1} B_k,j$, then $$\mu(E \setminus \bigcup_{k \geq 1} E_k) \leq \mu(\bigcup_{k \geq 1} (A_k \setminus E_k)) \leq \sum_{k \geq 1} \mu(A_k \setminus E_k)=0$$

The problem here is that I cannot affirm $f_n \rightrightarrows f$ on $E_k$.

Any suggestions to complete the solution would be greatly appreciated.

  • $\begingroup$ Convergence doesn't have to be uniform on the infinite union. If it was, Egorov's theorem could have been sharpened to assert there exists a single subset $B \subseteq E$ s.t. $\mu( E \setminus B ) = 0$ and $f_n \rightrightarrows f$ on $B$. $\endgroup$
    – Adayah
    Commented May 11, 2015 at 8:15
  • $\begingroup$ An also reasonable approach would be to take the union of $B_{k, j}$ over $k$ instead of $j$. But then again, the convergence wouldn't have to be uniform. $\endgroup$
    – Adayah
    Commented May 11, 2015 at 8:21

1 Answer 1


Hint: take $$E_j = \bigcup_{k=1}^j B_{k, j}.$$

  • $\begingroup$ Thanks for your answer. You are right, by taking finite union I can get uniform convergence in each $E_j$. I am having some trouble showing $\mu(E \setminus \bigcup_{j\geq 1} E_kj)=0$. I was trying to prove $\bigcup_{k \geq 1}A_k \setminus \bigcup_{j \geq 1}E_j \subset \bigcup_{k \geq 1}( A_k \setminus \bigcup_{j\geq 1}B_{k,j})$. It may be an extremely basic set theory problem but I was struggling with this inclusion, I don't know how to prove it. $\endgroup$
    – user16924
    Commented May 11, 2015 at 20:38
  • $\begingroup$ That's a good try, but has a little issue. Try proving this: $\displaystyle \bigcup_{k \geqslant 1} A_k \setminus \bigcup_{j \geqslant 1} E_j \subseteq \bigcup_{k \geqslant 1} \left( A_k \setminus \bigcup_{j \geqslant k} B_{k, j} \right). $ $\endgroup$
    – Adayah
    Commented May 12, 2015 at 5:05

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