Let $(\Omega,\mathcal{F},P)$ be a probability space, and $A\in\mathcal{F}$ such that $P(A)>0$. Let $P_{A}$ denote the conditional probability measure given $A$ defined on $(\Omega,\mathcal{F})$ Now obviously $dP_A/dP=1_A/P(A)$, so then for any $X\in L^1$
$$E[X|A]:=E_{P_A}[X]=\int XdP_A=\int X\frac{dP_A}{dP}dP=E[X1_A]/P(A)$$ Now let $\sigma(A)=\{\emptyset,A,\Omega\backslash A,\Omega\}$ denote the sigma filed generated by $A$. So we must have for any $\omega\in A$, $E[X|\sigma(A)](\omega)=E[X|A]1_A(\omega)$. Is that true? But I have trouble showing that that for any $\omega\in A$ and r.v. $X\in L^1$
$$E[X|\sigma(A)](\omega)=\frac{1}{P(A)}\int_\Omega X1_A(\omega)dP=\frac{1}{P(A)}\int_AXdP=E[X1_A(\omega)]/P(A):=E[X|A]$$
So I canot prove the above equation, though it seems quite obvious but am quite stuck or is it wrong?. I started with the definition of conditional expectation but am getting nowhere, the only thing I can see is, $E[X|\sigma(A)]=E[X|\sigma(A)]1_A+E[X|\sigma(A)]1_{\Omega\backslash A}$, and can't see where I can pull out $1/P(A)$, so any help will be greatly appreciated. So the question is in fact how are $E[X|\sigma(A)]$ and $E[X|A]$ related? If there are any mistakes or if I have got hings wrong please correct me. Thanks and any help feedback is most needed and appreciated.
Indeed, $E[X\mid\sigma(A)]=E[X\mid A]\,\mathbf 1_A+E[X\mid \Omega\setminus A]\,\mathbf 1_{\Omega\setminus A}$.