If $\{f_n\}$ is sequence of measurable functions on $X$, then $\{x: \lim f_n(x) \text{ exists}\}$ is a measurable set.

My idea is to prove that $$\{x: \lim f_n(x) \text{exists}\}=\{x: \liminf f_n(x)=a<\infty\}\cap \{x: \limsup f_n(x)=a<\infty\}=\bigg(\cup_{j=1}^{\infty}\cap_{n\geq j}\{x: f_n(x)=a\}\bigg)\cap\bigg(\cap_{j=1}^{\infty}\cup_{n\geq j}\{x: f_n(x)=a\}\bigg)$$ which two sets are measurable and their insection is also measurable. Is it correct?

Moreover, Notice that function $g=\limsup f_n-\liminf f_n$ is measurable because $f_n$ is measurable and by proposition 2.7. So $$\{x: \lim f_n\text{ exists} \}=\{x: \limsup f_n=\liminf f_n\}=\{x: g(x)=0\}.$$ is measurable which because $g^{-1}(\{0\})$ is measurable.

Another question: Do I need to consider that $ \limsup f_n=\pm \infty \text{ or } \liminf f_n=\pm \infty$

  • $\begingroup$ Your notation for intersections and unions is really confusing! How are these grouped? What's up with the union before the word "which"? $\endgroup$ – Milo Brandt Sep 26 '19 at 23:09
  • $\begingroup$ @KaviRamaMurthy I don't think that question is fully a duplicate - there are some misconceptions about sets and using them for measurability that are unique to this question, even if the other gets the same result - and I think that it's worth answering this one as its own question. $\endgroup$ – Milo Brandt Sep 26 '19 at 23:23

Suppose $f_n: X\to \mathbb{\bar{R}}$ for all $n\in \mathbb{N}$. If we can write $$ \{x\in X:\lim_{n\to\infty}f_n(x)\;\mathrm{exists}\}=E\cup E_{\infty}\cup E_{-\infty} $$ where $E_{\pm\infty}$ are the sets of $x$ such that the limit is equal to $\pm\infty$ and $E$ is the set of $x$ such that the limit exists. It is suffices to show that each term is measurable.

Notice that $\limsup f_n\text{ and} \liminf f_n$ are measurable because $f_n$ is measurable and by proposition 2.7. Define [g(x):= \limsup f_n(x)-\liminf f_n(x) ] where $\limsup f_n(x)=\liminf f_n(x) \notin\{\pm\infty\}$. So $g(x)$ is measurable. Thus, [E={x: \limsup f_n=\liminf f_n\notin{\pm\infty}}] implies the set $E$ can be written as $g^{-1}(\{0\})$ which is measurable.

Now $\lim_{n\to\infty}f_n(x)=\infty$ if for all $M\geq 1$ there exists $N$ such that $f_n(x)\geq M$ for all $n\geq N$, that is, $$ E_{\infty}=\bigcap_{M=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n\geq N}\{x:f_n(x)\geq M\}$$ hence is measurable. A similar argument works for $E_{-\infty}$, that is, $$ E_{\infty}=\bigcap_{M=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n\geq N}\{x:f_n(x)\leq -M\}$$

Suppose $f_n: X\to \mathbb{C}$ for all $n\in \mathbb{N}$. Since $f_n(x)$ converges if and only if $Re(f_n(x))$ and $Im(f_n(x))$ are both convergence, that is, $$\{x: \lim f_n\text{ exists}\}=\{x: \lim Re(f_n)\text{ exists}\}\cap \{x: \lim Im(f_n)\text{ exists}\}$$ which two sets are both measurable by the previous argument and their intersection is also true. This gives the desired result.


Here's another way of doing it. Define

$$ g:=\liminf_{n\to\infty} f_n $$

$$ h:=\limsup_{n\to\infty} f_n $$

The functions $g$ and $h$ are both measurable.

Then note that

$$ E:=\{x\in X: \lim_{n\to\infty} f_n(x) \text{ exists}\}=\{x\in X: g(x)=h(x)\} $$

As $g$ and $h$ are measurable, so is $E$.

Edit: If by "the limit exists" you mean also that it is finite, consider the set $$ A=\{ x\in X: \limsup_{n\to\infty} f_n(x) <\infty\} $$ Note that $A$ is measurable, and thus so is $E\cap A$, which is what you want.

  • $\begingroup$ But if $f_n: X\to \infty$, it does not work. $\endgroup$ – user469065 Sep 27 '19 at 4:08
  • $\begingroup$ @LoveQYG I edited my answer, does that answer your question? $\endgroup$ – Reveillark Sep 27 '19 at 4:10
  • $\begingroup$ The downvote seems rather shady. $\endgroup$ – Reveillark Sep 27 '19 at 4:14
  • $\begingroup$ Maybe $\lim f_n=\infty$ $\endgroup$ – user469065 Sep 27 '19 at 4:22
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    $\begingroup$ If you're looking at complex valued functions+, then they can't take the value $\infty$. As for the original problem, split each $f_n$ into real and imaginary parts. Then the sequence $f_n(x)$ converges if and only if both the real and imaginary parts of $f_n(x)$ converge, so $\{x: f_n(x)\text{ converges}\}$ is an intersection of two measurable sets, thus measurable. $\endgroup$ – Reveillark Sep 27 '19 at 18:42

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