I'm following Durrett's book Probability: Theory and Examples. On the proof of uniqueness of conditional expectation is used the following argument:

Suppose $Y$ and $Y'$ are two different conditional expectations for the random variable $X$ given $\mathcal{F}$, so $\int_A Y dP=\int_A Y' dP$ for all $A\in\mathcal{F}$. Taking $A=\{Y-Y'\geq \epsilon>0\}$ we see

$$0=\int_A Y-Y' dP\geq \epsilon P(A)$$

so $P(A)=0$, and this works for all $\epsilon>0$ so $Y\leq Y'$ a.s., we can proof $Y\geq Y'$ a.s. the same way, so the conditional expectation is unique up to random variables that are different on sets with probability $0$.

Now I need to use this argument to show that if $X_1=X_2$ on $B\in\mathcal{F}$, then $E(X_1|\mathcal{F})=E(X_2|\mathcal{F})$ a.s. on $B$.

Can anybody help me with that? Do I need to define a set similar to the $A$ on the initial argument?

  • $\begingroup$ Yes, and consider the intersection of the defined set with $B$ $\endgroup$ Nov 22, 2016 at 17:36
  • $\begingroup$ Thanks Petite, I used your suggestion and wrote an answer, can you please check if I am correct? $\endgroup$
    – Irene
    Nov 22, 2016 at 17:46
  • $\begingroup$ Yeah, roughly right, just notice that $B$ could be a strict subset of $\{X_1 = X_2\}$ and your argument works because the set $A$ defined in your answer is in $\mathcal{F}$ as both $Y_1$ and $Y_2$ are $\mathcal{F}$ measurable. $\endgroup$ Nov 22, 2016 at 18:35

1 Answer 1


Let $Y_1=E(X_1|\mathcal{F})$ and $Y_2=E(X_2|\mathcal{F})$. Consider $A=\{Y_1-Y_2\geq \epsilon>0\}$, so

$$0=\int_{A\cap B}X_1-X_2 dP=\int_{A\cap B}Y_1-Y_2 dP\geq\epsilon P(A\cap B)$$

then $P(A\cap B)=0$ and using the initial argument we have $Y_1=Y_2$ a.s. on $B$.


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