Sum of two closed sets are measurable

[hhj5575]

I tried very long time to show that
For closed subset A,B of R^d, A+B is measurable.

A little bit of hint says that it's better to show that A+B is F-simga set...
It seems also difficult for me as well...

Could you give some ideas for problems?

 

[Citan Uzuki]

Try showing that if A and B are compact, then A+B is also compact. Then in the general case where A and B are closed, can you see how to write A+B as the countable union of compact sets?

 

[hhj5575]

hmm.. I thought that if A,B is closed, then A+B is not necessarily closed...

 

[Citan Uzuki]

That's true, but you're not being asked to prove that A+B is closed, you're being asked to prove that it's a countable union of closed sets. That's why you start with the case where A+B is compact. If A and B are compact instead of just closed, then A+B really is compact. Then you can use the fact that every closed set is a countable union of compact sets to prove that A+B is F_σ

 

[hhj5575]

Aha. I got your idea!
I'm now wondering how I can show the fact that compactness is closed under the set addition..

Is it better to use open covering argument?

 

[Citan Uzuki]

Actually it's probably faster to use the fact that a subset of R^n is compact iff every sequence has a convergent subsequence.

 

[hhj5575]

hmm.. I'm not sure It is actually true that every seq in A+B has a convergent subseq.

assuring the exsitence of conv subseq in A and B does not implies so as well A+B... Isn't it?

 

[Citan Uzuki]

A sequence in A+B is of the form $\{a_n + b_n\}_{n\in \mathbb{N}}$, where $a_n\in A$ and $b_n\in B$. Can you find a subsequence such that $a_{n_k}$ converges and $b_{n_k}$ converges? What does that say about $a_{n_k} + b_{n_k}$?

 

[hhj5575]

I got it. Thank you very much!