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$F$ and $F'$ are $\sigma$-algebras, and $M$ is a function from $(\Omega,F)$ to $(\mathbb{R},B(\mathbb{R}))$

If this statement is true, how to reason or understand it in a simple way?

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    $\begingroup$ what are $M$, $F$ and $F'$? $\endgroup$
    – user251257
    Commented Jul 21, 2015 at 1:45
  • $\begingroup$ Just added more description. :) $\endgroup$
    – Cancan
    Commented Jul 21, 2015 at 1:49
  • $\begingroup$ it works the other way, if $F'$ measurable then also $F$ measurable. Think about it like how much information $M$ encodes. $\endgroup$
    – user251257
    Commented Jul 21, 2015 at 1:54
  • $\begingroup$ $F$-measurable means that the inverse image of any Borel set lies in $F$. If $F'$ is a proper subset of $F$ then obviously it is possible for a set to be in $F$ but not in $F'$. Just pick $F'$ to be some very small $\sigma$-algebra and it is trivial to find an example. $\endgroup$
    – Erick Wong
    Commented Jul 21, 2015 at 1:55

2 Answers 2

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The statement is false. For example, in the real line $\Bbb R$ it is possible to construct a Lebesgue-measurable set which is not Borel-measurable. This construction is non-trivial. You can check a lot of references in this thread.


Edit: the statement stays false, take $M$ as the identity in $\Bbb R$ and apply the above mentioned construction.

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    $\begingroup$ why so complicate, just pick $F' = \{\emptyset, \Omega\}$. $\endgroup$
    – user251257
    Commented Jul 21, 2015 at 1:51
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It is clear the answer is no! Because a measureable set must included in a $\sigma$-algebra then it need not be included in its subsets.

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