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Suppose that $(f_n)_n$ is a sequence of measurable functions on a set $E$ and that $f_n \to f$ a.e.on $E$. Does this imply that $f$ is measurable?

I know that pointwise limit of measurable function is measurable. But here we only have convergence a.e. So I got confused.

PhoemueX
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user145993
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2 Answers2

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That depends on the exact context.

In general, $f$ will not be measurable. To see this, simply take $(X, \{\emptyset, X\}, \mu)$ as your measure space, with $\mu(\emptyset) = 0 = \mu(X)$.

Then $f_n \to f$ almost everywhere holds for every sequence $(f_n)_n$ and every map $f$, but only constant maps are measurable.

Now assume that your measure space $(X, \Sigma, \mu)$ is complete. This means that if $A \in \Sigma$ with $B \subset A$ and $\mu(A) = 0$, then also $B \in \Sigma$.

Then $f$ is measurable. To see this, first note that if $f = g$ almost everywhere (i.e. on $N^c$ with $\mu(N) = 0$) and $g$ is measurable, then so is $f$, because for every interval $I \subset \Bbb{R}$, we have

\begin{eqnarray*} f^{-1}(I) &=& [f^{-1}(I) \cap N^c] \cup [f^{-1}(I) \cap N] \\ &=& [g^{-1}(I) \cap N^c] \cup [f^{-1}(I) \cap N]. \end{eqnarray*}

But $N,N^c$ are measurable and $g$ is measurable. Hence, $g^{-1}(I) \cap N^c$ is measurable.

Finally, since your measure space is complete, $f^{-1}(I) \cap N$ is measurable (because it is a subset of the null-set $N$).

Hence, $f^{-1}(I)$ is measurable.

Now let $g := \liminf_n f_n$. Then $g$ is measurable and $f = g$ almost everywhere because of $f_n \to f$ almost everywhere. By the above, this implies that $f$ is measurable.

Finally note that the Lebesgue measure equipped with the $\sigma$-algebra of Lebesgue measurable sets is complete, but equipped with the $\sigma$-algebra of Borel measurable sets, it is not complete.

PhoemueX
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  • I don't think your counterexample is quite right. Consider, for example, $X = {0,1}$, $f_n(x) = 0$ for all $n$, and $f(x) = x$. The set where the $f_n$ and $f$ disagree is ${1}$, which is not even measurable. Moreover, $f^{-1}(\emptyset) = \emptyset$ and $f^{-1}(X) = X$, so $f$ is a non-constant measurable function. – Tony Mottaz Sep 07 '16 at 02:49
  • @AnthonyMottaz: Ok, with the word "function" I was referring to functions $f: X \to \mathbb{R}$ or $f : X \to \mathbb{C}$. In this case, only constant functions are measurable. Also, it does not matter that ${1}$ is not measurable. The (at least my) definition of $f_n \to f$ a.e. is that there is some null-set $N$ (take $N=X$ in this setting) such that $f_n (x) \to f(x)$ for all $x \in X \setminus N$. But thanks for the comment. I will clarify my answer slightly this afternoon. – PhoemueX Sep 07 '16 at 05:10
  • Got it. That clears things up. Thanks. – Tony Mottaz Sep 07 '16 at 05:12
  • @PhoemueX. I added an alternative view on this - would appreciate your opinion. – Tom Collinge Mar 16 '18 at 11:39
  • @Sam: I have shown that if $f = g$ almost everywhere with $g$ measurable, then $f$ is measurable as well. In the final part, I then construct the suitable (measurable) $g$ to apply this observation. – PhoemueX Dec 02 '20 at 10:49
  • @PhoemueX if we need to show $f$ is not measurable a.e. we need to take $X\setminus N$ is not measurable. Right? – Aritra Jun 30 '21 at 15:21
  • How do you define "almost everywhere"? According to my lecture notes, this only means that $$\inf\left{\mu(A):N\subset A\text{ and }A\in\Sigma\right}=0$$and not necessarely that $N$ is measurable. In this case, your argument does not work does it? I have already asked this question here, maybe you can have a look. – Filippo Jan 22 '22 at 11:28
  • @Filippo: My definition of "property $A(x)$ holds for almost all x" is that there is a (measurable) set N with $\mu(N)=0$ so that $A(x)$ is true for all $x \notin N$. In your case, note by taking $A_n \in \Sigma$ with $\mu(A_n)\leq 1/n$ and $A_n \supset N$ and then taking $N' = \bigcap_n A_n$ that there always exists a measurable(!) null-set $N'\supset N$ with $\mu(N')=0$. Thus, our definitions/interpretations should be identical in the end. – PhoemueX Jan 22 '22 at 14:07
  • @PhoemueX Thank you very much for the reply :) This does actually solve my problem, because I know that there is a measurable function $f$ on $X$ such that $f=\lim_{n\to\infty}f_n$ on $X\setminus N$. – Filippo Jan 24 '22 at 16:19
  • I have a strongly related question in which the completeness of a measure space is the crucial point. Could you please have a look at my question? Thank you so much for your help! – Akira Aug 07 '22 at 15:23
  • @PhoemueX Would a counterexample exist if we restrict to probability spaces? I ask this because we routinely take almost sure limits to be measurable but we don't necessarily require completeness on probability spaces. – mathnoob123 Nov 12 '22 at 06:08
  • @mathnoob123: A counterexample exists for every non-complete measure space. Take a null set $N$ and a non-measurable subset $N'\subset N$ and $f_n=0$, $f=1_{N'}$. Then you have convergence almost everywhere, but the limit is not measurable. – PhoemueX Nov 12 '22 at 07:55
  • If I've understood it correctly, we've been talking about a.e. convergence, but what happens if we consider $p$-convergence (i.e. $\mu$-convergence)? So if $f_{n}\overset{\mu}{\rightarrow}f$ (in the sense that $\lim_{n\rightarrow\infty}\mu\left(x:\left|f_{n}\left(x\right)-f\left(x\right)\right|>\varepsilon\right)=0$ for any arbitrary $\varepsilon)$, then is it true that the $p$-limit $f$ of the measurable functions $f_{n}$ is still measurable? Is there any additional assumption next to the completness of the underlying space that we have to assume in order to make the mentioned statement true? – Kapes Mate Nov 10 '23 at 10:20
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    @KapesMate: First, it is difficult to define convergence in measure without knowing that the limit is measurable (you won't know that $\mu ( {x : |f_n(x) - f(x)| > \varepsilon})$ makes sense if you don't know that $f$ is measurable). Also, if you have convergence in measure, then a subsequence converges almost everywhere (https://math.stackexchange.com/questions/1006091/convergence-in-measure-implies-convergence-almost-everywhere-of-a-subsequence), but it might well be that the proof already uses that $f$ is measurable. – PhoemueX Nov 10 '23 at 12:26
  • Fair point. So when you “calculate” $\mu\left(x:\left|f_{n}\left(x\right)-f\left(x\right)\right|>\varepsilon\right)$, then you implicitly assume that $f$ is measurable, because otherwise it doesn't make sense. Thanks. – Kapes Mate Nov 10 '23 at 12:40
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I agree with the answer by @PhoemueX, but maybe not with the question ! Consider an alternative .......

Suppose that $\{f_n\}$ is a $\mu$-.a.e. convergent sequence and define $f = \lim_{n \rightarrow \infty} f_n$ where the limit exists. Then $f$ is a $\mu$-.a.e. defined function and $f$ is measurable.

Clearly $f$ is $\mu$-.a.e. defined, to prove measurability.....

... let A be the null-set on which $\{f_n\}$ fails to converge.
Then $f_n |_{X \setminus A}$ is a measurable function on $X \setminus A$ and converges to a limit $f$ on $X \setminus A$ which is measurable by the normal convergence theorem. Then $f$ is a $\mu$-.a.e. defined measurable function on X.

In terms of the counterexample in the previous answer, only constant functions are measurable and therefore a sequence of constants converge everywhere or nowhere. In the latter case, $X$ is the "null-set" where the sequence fails to converge and $f$ is defined "almost everywhere" except on $X$ - i.e. it is nowhere defined, but this still fits the definition of $\mu$-.a.e. defined in this case.

Tom Collinge
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    This looks good; just two points: 1) If the underlying measure space is not complete, then talking about "a.e. defined functions" is a bit tricky. Maybe it would be better to just set $f(x) = 0$ for $x \in A$. With this choice, we definitely have measurability. 2) I would maybe add a brief remark on why the set $X \setminus A$ on which pointwise convergence occurs is measurable. This is because $X \setminus A = \bigcap_{n=1}^\infty \bigcup_{m=1}^\infty \bigcap_{k,\ell = m}^\infty {x \in X ,:, |f_k (x) - f_\ell(x)| \leq 1/n }$. – PhoemueX Mar 18 '18 at 08:57
  • @PhoemueX Can you please explain further why $X-A=\bigcap_{n=1}^{\infty}\bigcup_{m=1}^{\infty}\bigcap_{k,l=m}^{\infty}{x\in X:|f_k(x)-f_l(x)|\leq 1/n}$ implies the measurability of $f$? – Gregoire Rocheteau Mar 19 '18 at 02:58
  • @Ya: Because then $f_n \cdot 1_{X\setminus A}$ converges to $f$ (which we set to zero on $A$) everywhere, and these are measurable(!), and the pointwise limit of measurable functions is measurable. – PhoemueX Mar 19 '18 at 07:18
  • @PhoemueX Can you explain why your first point (1) holds, that is if we set $f(x)=0$ for $x\in A$, then $f$ is measurable? – nan Apr 10 '18 at 15:30
  • @nan: Because $f_n \cdot 1_{A^c}$ is measurable, and $f_n \cdot 1_{A^c} \to f$ pointwise. – PhoemueX Apr 11 '18 at 21:19