Set of points at which sequence of measurable functions converge (another approach)

Question is to prove that :

Set of all points at which a sequence of measurable functions converge is a measurable set..

What i have tried is as follows :

We are looking at the following set :

$$\{x\in X : (f_n(x)) \text{converges}\}$$

Which is same as $$\{x\in X : (f_n(x)) \text{is cauchy}\}$$

Which is same as

$$\{x\in X : \text{given $\epsilon>0$ there exists $N\in \mathbb{N}$ such that $|f_n(x)-f_m(x)|<\epsilon$ for all $m,n\geq N$}\}$$

I want to write this as unions and intersections of measurable sets and then conlcude this is measurable.. Something like :

$$\bigcap_{p\in \mathbb{R}??}\bigcup_{m,n\in \mathbb{N}??}\{x:|(f_n-f_m)(x)|<p\}$$

Now, As $f_n,f_m$ are measurable so is $f_n-f_m$ and so is $|f_n-f_m|$ and $\{x:|(f_n-f_m)(x)|<p\}$ being inverse image of open set is measurable..

I am not so sure how to write that as unions and intersections...

Solution 1:

$(f_n(x))_n$ is cauchy means that for any positive integer $p$, there exists one integer $k$, such that for all $m > k$ and $n > k$ we have $|f_n(x) - f_m(x)| < \frac{1}{p}$.

Since it's for any positive integer p, we should have something like $\cap_{p=1}^{\infty}$.

And there exists a $k$ such that blabla means $\cup_{k=1}^{\infty}$ blabla..., i.e. for one k in $\{1,2, \cdots\}$, blabla is ok

For all $m,n$ greater than $k$ is $\cap_{m > k, n>k}$.

So finally $(f_n(x))_n$ is cauchy means that $x$ is in the set

$\cap_{p=1}^{\infty} \cup_{k=1}^{\infty}\cap_{m > k, n>k}\{x: |f_m(x) - f_n(x)| < \frac{1}{p}\}$

To resume, when there is "for any, for all", use intersection; when there is "exists", use union

Solution 2:

The idea is good. But use a different definition of continuity which is in fact equivalent: Namely $$\{x\in X : \text{given $\mathbb N\ni k>0$ there exists $N\in \mathbb{N}$ such that $|f_n(x)-f_m(x)|<\frac{1}{k}$ for all $m,n\geq N$}\}$$

Then write this as

$$\bigcap_{k\in \mathbb{N}}\bigcup_{N\in\mathbb N}\bigcap_{m,n\geqslant N}\left\{ x:\left|(f_n-f_m)(x)\right|<\frac{1}{k}\right\}$$

As you said the set $\left\{ x:\left|(f_n-f_m)(x)\right|<\frac{1}{k}\right\}$ is measurable and hence you're done, since you have a countable union respectively intersection of measurable sets. This is why you use the different definition of continuity.