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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.

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    $\begingroup$ As in the very recent question that (you probably deleted and) this one is a duplicate of, "E[X|σ(A)](ω)=E[X|A]1A(ω)" is pure fantasy, as well as the displayed formula right after it. (No need to ask a third question, these inconvenient facts will remain the same.) Do you have a definition of E[X|G] at your disposal, when G is a sigma-algebra? $\endgroup$ – Did May 23 '14 at 11:03
  • $\begingroup$ First of all I may not of put things across correctly, so confusion on my part as well as yours might of arisen, my apologies. I was trying to verify a comment in Koralav and Sinai and what I am trying to say is that on $A$, we have $E[X|\sigma(A)]=E[X|A]$. I now see this is (trivially)true as on $A$ $$\int_A E[X|\sigma(A)]dP= \int_A XdP=E[X1_A]=(E[X1_A]/P(A))\int_A1dP=\int_AE[X|A]dP$$. In fact $E[X|\sigma(A)]=E[X|A]1_A+E[X|\Omega\backslash A]1_{\Omega\backslash A}$. $\endgroup$ – user152874 May 23 '14 at 14:53
  • $\begingroup$ Furthermore if $\mathcal{G}$ is a sub sigma field of $\mathcal{F}$ generated by a countable partition $\{G_i\}_{i=0}^{\infty}$ with $P(G_0)=0$ and $P(G_i)>0$ $\forall i=1,2,...$ then $E[X|\mathcal{G}]=\sum\limits_{i=1}^{\infty} E[X|G_i]1_{G_i}$ $\endgroup$ – user152874 May 23 '14 at 14:59
  • $\begingroup$ Which confusion "on my part"? Curious statement... // Does the answer below allow to close this discussion? $\endgroup$ – Did May 23 '14 at 18:13
  • $\begingroup$ Sorry, contrarily to what I suggested above, you did not delete the previous question, you probably logged under a new username to repost the same one. $\endgroup$ – Did May 23 '14 at 23:08
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Indeed, $E[X\mid\sigma(A)]=E[X\mid A]\,\mathbf 1_A+E[X\mid \Omega\setminus A]\,\mathbf 1_{\Omega\setminus A}$.

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