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

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 $$\limsup f_n=\pm \infty \text{ or } \liminf f_n=\pm \infty$$

• 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

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.

• But if $f_n: X\to \infty$, it does not work. – user469065 Sep 27 '19 at 4:08
• @LoveQYG I edited my answer, does that answer your question? – Reveillark Sep 27 '19 at 4:10
• The downvote seems rather shady. – Reveillark Sep 27 '19 at 4:14
• Maybe $\lim f_n=\infty$ – user469065 Sep 27 '19 at 4:22
• 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