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Let $(X,\Sigma,\mu)$ be a measure space and suppose $f_n\colon X\to\mathbb{R}$ is measurable for all $n\in\mathbb N$. It is well known that if $$\lim_{n\to\infty}f_n(x)$$ exists for all $x\in X$, then $$\left(\lim_{n\to\infty}f_n\right)(x):=\lim_{n\to\infty}f_n(x)$$ is a measurable function. But what if we only know that the limit exists almost everywhere? Note:

  • The set $N$ on which the limit does not exist is not necessarely measurable - at least according to the definition of "almost everywhere" from my lecture.$^1$

  • I am not asking whether the function $$\lim_{n\to\infty}f_n\colon X\setminus N\to\mathbb{R}$$ is measurable, but if the function \begin{align}\lim_{n\to\infty}f_n\colon X&\to\mathbb{R}\\x&\mapsto\begin{cases}\lim_{n\to\infty}f_n(x)&\text{if }x\in X\setminus N\\0&\text{else}\end{cases}\end{align} is measurable (or are these equivalent?).

$^1$ According to the definition from my lecture this only tells us that $$\inf\left\{\mu(A):N\subset A\text{ and }A\in\Sigma\right\}=0$$

Filippo
  • 3,536
  • See this. – David Mitra Jan 22 '22 at 10:51
  • @DavidMitra Thank you, but the accepted answer seems to assume that $N$ is measurable and the second answer seems to answer another question (my question was a bit ambiguous, so I edited it). – Filippo Jan 22 '22 at 11:14
  • I think completeness of the measure is what's essential. Here is a counterexample from Hewitt-Stromberg's Real and Abstract Analysis: Let $X=\Bbb R$, the measurable sets be the Borel sets, and $\mu$ be Lebesgue measure. Let $A$ be a non-measurable subset of the Cantor set. Set each $f_n$ to be identically 0 and $f$ to be the indicator function on $A$. (Somehow I thought the first answer in the link I gave above made this point.) – David Mitra Jan 22 '22 at 11:51
  • @DavidMitra Thank you for the example. PhoemueX explained to me that we can essentially assume that $N$ is measurable WLOG. This solves my problem :) – Filippo Jan 24 '22 at 16:22

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