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Let's $\mu$ a probability measure on $(\Omega, \mathcal{F})$. Let's $\mathcal{J}_1\supset\dots\supset\mathcal{J}_n\supset\dots$ a sequence of sub-$\sigma$-fields of $\mathcal{F}$. If $A\in\mathcal{J}_n$ for all $n\in\mathbb{N}$ is well known that $$ \mu(A|\mathcal{J}_n)=1_{A} $$ Here $1_{A}$ is the caracteristic of $A$.

Question. How to prove that if $A\in\displaystyle\bigcap_{n\in\mathbb{N}}\mathcal{J}_n$ then
$$ \mu(A|\displaystyle\bigcap_{n\in\mathbb{N}}\mathcal{J}_n)=1_A\;? $$

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    What happens if you apply the definition of conditional probability?2012-03-31
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    When you say "is well known that," that is a definition from considering $E(1_A | \mathcal{F}_n)$ which equals $1_A$ by the rules of conditional expectation. To show what you want in your question, you simply need to verify that the rules of conditional expectation are satisfied, i.e. the integrals over all sets $B$ in the intersection work out to give $\int_B E(1_A | \cap_n \mathcal{F}_n)d\mu=\int_B 1_A d\mu$, in that $1_A$ is measurable w.r.t. the intersection2012-03-31
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    @GEdgar, Thank's.2012-03-31
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    Notice that $\mathcal{J} = \bigcap_{n \in \mathbb{N}} \mathcal{J}_n$ is a sub-$\sigma$-field. By your first statement, it follows that it is well known that $\mu(A|\mathcal{J}) = 1_A$.2012-04-12

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We have to use and show the following:

If $\mathcal A\subset\mathcal F$ is a $\sigma$-algebra and $A\in\mathcal A$ then $\mu(A\mid\mathcal A):=E[\mathbb 1_A\mid\mathcal A]=\mathbb 1_A$. (we can write the condition expectation as $\mathbb 1_A$ is bounded, hence integrable)

As $\mathbb 1_A$ is $\mathcal A$-measurable, it's the best candidate to be a version of its conditional expectation.