How can I rewrite this sets using measure theory?

Question

Let $(\Omega, F, \mu)$ be a measure space. Let $f_n:\Omega\rightarrow \Bbb{R}$ be measurable functions. Let $$B:=\{x\in \Omega: \lim_{n\rightarrow \infty} f_n(x)~~\text{does not exists}\}$$I need to rewrite this in terms of limit superior and limit inferior of sets in $F$.

My idea was the following:

We know that $\lim_{n\rightarrow \infty} f_n(x)$ does not exists iff $$\exists \epsilon>0:~~\forall~n\geq 1~~\exists~k\geq n~~s.t.~~|f_k(x)|>\epsilon$$ Therefore I thought that we can write $$B=\bigcup_{\epsilon>0}\bigcap_{n\geq 1}\bigcup_{k\geq n} \{|f_k(x)|>\epsilon\}$$ But now we still need to rewrite the set $\{|f_k(x)|>\epsilon\}$. I would say that this is equal to $A_k:=f_k^{-1}((\epsilon,\infty))\cup f_k^{-1}((-\infty, -\epsilon))$. Then this union is clearly in $F$ since $f_k$ are measurable. Then $$B=\bigcup_{\epsilon>0}\bigcap_{n\geq 1}\bigcup_{k\geq n}A_k=\bigcup_{\epsilon>0}\limsup_{n\rightarrow \infty}A_n$$

Does this work like this?

Thanks for your help.

Answer 1

I assume you would like to show that the set of points of nonconvergence is measurable. Your condition is incorrect, since it means that $f_n(x)\not\to 0$. These are my two tries.

First approach

• $\lim_{n\rightarrow \infty} f_n(x)$ exists iff $\exists a\in \Bbb R\,\forall \varepsilon>0\,\exists N\,\forall n\geq N:|f_n(x)-a|<\varepsilon$.
• Therefore $\lim_{n\rightarrow \infty} f_n(x)$ does not exist iff $\forall a\in\Bbb R\,\exists \varepsilon>0\,\forall N\,\exists n\geq N:|f_n(x)-a|\geq \varepsilon$
• So $\lim_{n\rightarrow \infty} f_n(x)$ does not exist iff $$x\in A:=\bigcap_{a\in\Bbb R}\bigcup_{\epsilon>0}\bigcap_{N\in\Bbb N}\bigcup_{n\geq N}\{x:|f_n(x)-a|\geq \varepsilon\} \\ = \bigcap_{a\in\Bbb R}\bigcup_{k\in\Bbb N}\bigcap_{N\in\Bbb N}\bigcup_{n\geq N}\left\{x:|f_n(x)-a|\geq \frac 1k\right\} \\ =\bigcap_{a\in\Bbb R}\bigcup_{k\in\Bbb N}\bigcap_{N\in\Bbb N}\bigcup_{n\geq N}F_{n}\left(a,\frac 1k\right) = \bigcap_{a\in\Bbb R}\bigcup_{k\in\Bbb N}\limsup_{n\to\infty} F_{n}\left(a,\frac 1k\right) $$

This description doesn't show that this set is measurable, infortunately, so instead of using the definition ofthe limit where one has to use the index $a$ from the uncountable set it's sometimes better to use the Cauchy condition.

Second approach

• $\lim_{n\rightarrow \infty} f_n(x)$ exists iff $\forall \varepsilon>0\,\exists N\,\forall m,n\geq N:|f_n(x)-f_m(x)|<\varepsilon$.
• Therefore $\lim_{n\rightarrow \infty} f_n(x)$ does not exist iff $\exists \varepsilon>0\,\forall N\,\exists m,n\geq N:|f_n(x)-f_m(x)|>\varepsilon$ (yes, we can have strict inequality).
• So $\lim_{n\rightarrow \infty} f_n(x)$ does not exist iff $$x\in B:=\bigcup_{\epsilon>0}\bigcap_{N\in\Bbb N}\bigcup_{m,n\geq N}\{x:|f_n(x)-f_m(x)|>\varepsilon\} \\ = \bigcup_{k\in\Bbb N}\bigcap_{N\in\Bbb N}\bigcup_{m,n\geq N}\left\{x:|f_n(x)-f_m(x)|>\frac 1k\right\} \\ = \bigcup_{k\in\Bbb N}\bigcap_{N\in\Bbb N}\bigcup_{m,n\geq N} F_{m,n}\left(\frac 1k\right)$$
• The sets $F_{m,n}(\frac 1k)$ are measurable, since $$F_{m,n}\left(\frac 1k\right)=\left\{x:|f_n(x)-f_m(x)|>\frac 1k\right\}=\\\bigcup_{q\in\Bbb Q}\{x: f_n(x)>q\}\cap\left\{x:f_m(x)<q-\frac 1k\right\}\cup \{x: f_m(x)>q\}\cap\left\{x:f_n(x)<q-\frac 1k\right\}$$