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This is probably a trivial question, but hopefully someone can help me with it:

Let $(\Omega,\mathcal{F},\mu)$ be a probability space, i.e., $\mathcal{F}$ is a $\sigma$-field on $\Omega$ and $\mu$ is probability measure on $\mathcal{F}$ (in particular, $\mu$ is countably additive, and $\mu(\Omega)=1$). Let $\mathcal{F}'$ be a $\sigma$-field on $\Omega$ such that $\mathcal{F}' \supsetneq \mathcal{F}$.

What are the necessary and/or sufficient conditions under which the following is true:

There exists a probability measure $\mu'$ on $\mathcal{F}'$ such that for each $A \in \mathcal{F}$, $\mu'(A) = \mu(A)$.

I know that it is possible to construct such a probability measure $\mu'$ if $\mathcal{F}$ has atoms -- but does there exist such a $\mu'$ if the $\sigma$-algebras do not have atoms?

Thanks!

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Think about $\mathcal F' = 2^\Omega$. –  cardinal Mar 4 '13 at 3:03
    
@cardinal: Thanks! Do you mean to say that any probability measure $\mu$ on some $\sigma$-algebra on $\Omega$ can be extended to some probability measure $\mu'$ on the power set? I suspected that, but I can't see the formal argument. And if the above is true, can we then obtain a probability measure $\mu''$ on any sub-$\sigma$ algebra $\mathcal{F}''$ of the power set by restricting $\mu'$ to $\mathcal{F}''$? Thanks for your help -- I know I must be missing something trivial. –  Meeuw Mar 4 '13 at 15:16
    
Related: math.stackexchange.com/q/200677/7003 –  cardinal Mar 5 '13 at 3:30
    
@cardinal - many thanks! I should have seen this right away. For the record/future readers: The answer is: This cannot be done in general. For example, the Lebesgue measure on (the Borel $\sigma$-algebra on) [0,1] cannot be extended to the power set. –  Meeuw Mar 5 '13 at 18:10
    
Assuming the continuum hypothesis. –  Did Mar 22 '13 at 8:15

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