# If $M$ is $F$-measurable, then is it also $F'$-measurable with $F'\subset F$?

$F$ 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?

• what are $M$, $F$ and $F'$? – user251257 Jul 21 '15 at 1:45
• Just added more description. :) – Cancan Jul 21 '15 at 1:49
• it works the other way, if $F'$ measurable then also $F$ measurable. Think about it like how much information $M$ encodes. – user251257 Jul 21 '15 at 1:54
• $F$-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. – Erick Wong Jul 21 '15 at 1:55

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.
• why so complicate, just pick $F' = \{\emptyset, \Omega\}$. – user251257 Jul 21 '15 at 1:51
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.