Sum of two closed sets are measurable

In summary, it seems that it is difficult to show that A+B is measurable. It is also difficult to show that compactness is closed under set addition. However, it is possible to use open covering argument to show that A+B is compact.
  • #1
hhj5575
8
0
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?
 
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  • #2
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?
 
  • #3
hmm.. I thought that if A,B is closed, then A+B is not necessarily closed...
 
  • #4
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
 
  • #5
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?
 
  • #6
Actually it's probably faster to use the fact that a subset of R^n is compact iff every sequence has a convergent subsequence.
 
  • #7
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?
 
  • #8
A sequence in A+B is of the form [itex]\{a_n + b_n\}_{n\in \mathbb{N}}[/itex], where [itex]a_n\in A[/itex] and [itex]b_n\in B[/itex]. Can you find a subsequence such that [itex]a_{n_k}[/itex] converges and [itex]b_{n_k}[/itex] converges? What does that say about [itex]a_{n_k} + b_{n_k}[/itex]?
 
  • #9
I got it. Thank you very much!
 

What does it mean for a set to be measurable?

A measurable set is one that can be assigned a numerical value, known as its "measure," according to a specific set of rules or criteria. In mathematics, this typically refers to the Lebesgue measure, which is a generalization of the concept of length, area, and volume.

What is the definition of a closed set?

A closed set is a set that contains all of its limit points. This means that if you were to draw a circle around any point in the set, the entire circle would be contained within the set. In other words, there are no points on the boundary of the set that are not included in the set itself.

What is the sum of two closed sets?

The sum of two closed sets is the set of all possible combinations of elements from both sets. In other words, it is the set of all elements that can be formed by adding an element from one set to an element from the other set. This operation is also known as the union of two sets.

Why is the sum of two closed sets important?

The sum of two closed sets is important because it allows us to combine two sets into one larger set. This can be useful in various mathematical and scientific applications, such as calculating probabilities, defining boundaries, and solving optimization problems.

How do you prove that the sum of two closed sets is measurable?

In order to prove that the sum of two closed sets is measurable, we can use the fact that the Lebesgue measure is countably additive. This means that the measure of the union of countably many disjoint sets is equal to the sum of their individual measures. By applying this property, we can show that the sum of two closed sets is indeed measurable.

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