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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.

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  • $\begingroup$ What are you confused about? Is the underlying measure space complete? $\endgroup$
    – copper.hat
    Jan 7, 2015 at 21:09

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$ Sep 7, 2016 at 2:49
  • $\begingroup$ @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. $\endgroup$
    – PhoemueX
    Sep 7, 2016 at 5:10
  • $\begingroup$ Got it. That clears things up. Thanks. $\endgroup$ Sep 7, 2016 at 5:12
  • $\begingroup$ @PhoemueX. I added an alternative view on this - would appreciate your opinion. $\endgroup$ Mar 16, 2018 at 11:39
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    $\begingroup$ @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 (math.stackexchange.com/questions/1006091/…), but it might well be that the proof already uses that $f$ is measurable. $\endgroup$
    – PhoemueX
    Nov 10, 2023 at 12:26
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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.

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    $\begingroup$ 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 $X \setminus A = \bigcap_{n=1}^\infty \bigcup_{m=1}^\infty \bigcap_{k,\ell = m}^\infty \{x \in X \,:\, |f_k (x) - f_\ell(x)| \leq 1/n \}$. $\endgroup$
    – PhoemueX
    Mar 18, 2018 at 8:57
  • $\begingroup$ @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 $f$? $\endgroup$ Mar 19, 2018 at 2:58
  • $\begingroup$ @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. $\endgroup$
    – PhoemueX
    Mar 19, 2018 at 7:18
  • $\begingroup$ @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? $\endgroup$
    – nan
    Apr 10, 2018 at 15:30
  • $\begingroup$ @nan: Because $f_n \cdot 1_{A^c}$ is measurable, and $f_n \cdot 1_{A^c} \to f$ pointwise. $\endgroup$
    – PhoemueX
    Apr 11, 2018 at 21:19

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