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...


$(f_n(x))_n$ 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

  • $\begingroup$ i understand this partially... please elaborate it a bit... I usually get stuck at this point only... $\endgroup$ – user87543 Jul 30 '14 at 15:10
  • $\begingroup$ You mean which point that is still not clear? $\endgroup$ – Petite Etincelle Jul 30 '14 at 15:15
  • $\begingroup$ now it is fine.. $\endgroup$ – user87543 Jul 31 '14 at 5:41
  • $\begingroup$ Thanks a ton times for the last line! $\endgroup$ – Error 404 Mar 29 '18 at 17:31

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.

  • $\begingroup$ I do not understand how do you write that as unions and intersections... please explain that part... $\endgroup$ – user87543 Jul 30 '14 at 15:08
  • 1
    $\begingroup$ Say an element $x$ is in $A_n$ for all $n$, then $x\in \bigcap _n A_n$ by unwinding the definition. Similarly, if there exists $n$ such that $x\in A_n$ then $x\in \bigcup_n A_n$. It is only another way of writing things down. $\endgroup$ – frog Jul 30 '14 at 15:12

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