- #1

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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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- Thread starter hhj5575
- Start date

- #1

- 8

- 0

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?

- #2

- 299

- 20

- #3

- 8

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

- #4

- 299

- 20

- #5

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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

- 299

- 20

- #7

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

- #8

- 299

- 20

- #9

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I got it. Thank you very much!

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