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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$

Daniel Fischer
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  • Your notation for intersections and unions is really confusing! How are these grouped? What's up with the union before the word "which"? – Milo Brandt Sep 26 '19 at 23:09
  • @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. – Milo Brandt Sep 26 '19 at 23:23

2 Answers2

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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.

Reveillark
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  • But if $f_n: X\to \infty$, it does not work. –  Sep 27 '19 at 04:08
  • @LoveQYG I edited my answer, does that answer your question? – Reveillark Sep 27 '19 at 04:10
  • The downvote seems rather shady. – Reveillark Sep 27 '19 at 04:14
  • Maybe $\lim f_n=\infty$ –  Sep 27 '19 at 04:22
  • Then your set is empty, in particular measurable. That’s not a problem – Reveillark Sep 27 '19 at 04:23
  • I mean $f_x: X\to \mathbb{\bar{R}}=\mathbb{R}\cup{{\infty}}$ –  Sep 27 '19 at 04:36
  • I assume you meant $f_n$ in place of $f_x$. And it’s still not a problem. – Reveillark Sep 27 '19 at 04:37
  • I feel like $\lim f_n=\infty$ is also the limit exists. –  Sep 27 '19 at 17:32
  • I don’t understand what your objection is. If you’re allowing the limit to be infinite, then the set of points of convergence is what I called $E$, and it is measurable. If you want the limit to additionally be finite, then you intersect with $A$, which preserves measurability. – Reveillark Sep 27 '19 at 17:50
  • Yes, so I feel like that it needs to discuss several situations including $f_n: X\to \mathbb{C} \text{or} \mathbb{\bar{R}}$ –  Sep 27 '19 at 18:40
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    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. – Reveillark Sep 27 '19 at 18:42
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Suppose $f_n:X\to\overline{\Bbb R}$ for all $n\in\Bbb N.$

We can write $$\left\{x\in X:\lim_{n\to\infty}f_n(x)\text{ exists}\right\}=E\cup E_\infty\cup E_{-\infty},$$ where $E_{\pm\infty}:=\left\{x\in X\mid\lim\limits_{n\to\infty}f_n(x)=\pm\infty\right\}$ and $E:=\left\{x\in X\mid\lim\limits_{n\to\infty}f_n(x)\text{ exists and is finite}\right\}.$

It 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, $g$ is measurable.

Thus, $E=\left\{x: \limsup f_n=\liminf f_n\notin\{\pm\infty\}\right\}$ implies the set $E$ can be written as $g^{-1}(\{0\}),$ which is measurable.

Now $\lim\limits_{n\to\infty}f_n(x)=\infty$ 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 $\operatorname{Re}(f_n(x))$ and $\operatorname{Im}(f_n(x))$ are both convergent, that is, $$\left\{x\in X:\lim_{n\to\infty} f_n\text{ exists}\right\}=\left\{x: \lim_{n\to\infty}\operatorname{Re}(f_n)\text{ exists}\right\}\cap\left\{x\in X:\lim_{n\to\infty}\operatorname{Im}(f_n)\text{ exists}\right\}$$ which two sets are both measurable by the previous argument and their intersection is also true. This gives the desired result.

PinkyWay
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