[944] Question 4 - HW1

[David Itkin]

Hi Lucas,

This argument shows that the set R (i.e. Omega) is in your sigma-algebra
(which is always true -- the entire space is always an element of the sigma
algebra), but that does not mean all subsets will be. Of course if you take
the power set of R as your sigma algebra then you will have all the
subsets, but this will not be true for other sigma algebras, in particular
the Borel sigma algebra B.

Actually constructing a non-Borel set is not so simple, but wikipedia has a
brief discussion on it:
https://en.wikipedia.org/wiki/Borel_set#Non-Borel_sets if you are curious.

Hope this helps,

David

On Sat, Oct 28, 2017 at 11:38 AM, Lucas Duarte Bahia <
lduarteb at andrew.cmu.edu> wrote:

> Why Sigma Algebra does not contain all subsets of R?
>
> Please point the flaw in the following logic:
>
> Capital Omega = R
> ]-infinity, alpha] complement = ]alpha, +infinity[
>
> Sigma Algebra contains  ]-infinity, alpha] and its complement, so it
> should contain Capital Omega = R
>
> Best Regards,
> *Lucas Duarte Bahia*
> MS. Computational Finance Student
> Carnegie Mellon University
> Telephone: (412) 378-1892
>
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