I have a question on the proof Durrett (p. $190$) gives for the uniqueness of the conditional expectation function.

If I understand his proof correctly, here is what I think it is saying:

Suppose $Y, Y'$ both satisfy the criteria to be a conditional expectation function for a random variable $X$ given a $\sigma$-algebra $\mathcal{A}$. Then if $A_{\epsilon} := \{ \omega \mid Y - Y' \geq \epsilon \}$, clearly $A_{\epsilon} \in \mathcal{A}$, and so by the criteria we have for conditional expectation, we have:

$$0 = \int \limits_{A_{\epsilon}} Y - Y' \,dP \geq \epsilon P(A_{\epsilon}) \geq 0$$ and this shows that $P(A_{\epsilon}) = 0$ for all $\epsilon > 0$.

Ok, then we note that $0 \leq P( \{\omega \mid Y - Y' > 0 \} ) = P(\bigcup \limits_{n =1}^{\infty} A_{\frac{1}{n}}) \leq \sum \limits_{n = 1}^{\infty} P(A_{\frac{1}{n}}) = \sum \limits_{n = 1}^{\infty} 0 = 0$. Thus, $P( \{\omega \mid Y - Y' \leq 0 \} ) = 1$, so $Y \leq Y'$ almost surely.

We can repeat the same proof switching $Y$ and $Y'$ to get $Y' \leq Y$ almost surely, thus they are equal almost surely.

My question:

I think the following proof is a much simpler one (using the fact that $\int \limits_{\Omega} X \,dP = 0 \implies X = 0$ almost surely). Is there any flaw in the argument below? If not, why did Durrett decide to do so many extra steps?

Since $\Omega \in \mathcal{A}$, we have $0 = \int \limits_{\Omega} Y - Y' \,dP$, and thus $Y - Y' = 0$ almost surely, so $Y = Y'$ a.s.


1 Answer 1


The problem with your argument is that $\int_{\Omega} X\,dP = 0$ does not imply that $X=0$ almost surely, unless you already know that $X \geq 0$ almost surely. Since you don't know that $Y-Y' \geq 0$ in your argument, your conclusion does not follow.

  • $\begingroup$ I am ridiculous for forgetting that it only holds when $X \geq 0$ a.s. Thank you for your sharp eye! $\endgroup$
    – layman
    May 13, 2015 at 18:58
  • $\begingroup$ Why is the limiting argument desirable? It should be enough to consider the events $A_1 = \{ Y \geq Y' \}$ and $A_2 = \{ Y \leq Y' \}$. We would have $Y_1 = Y_2$ a.s. on $A_1 \cup A_2$. The result follows since $\Omega \subset A_1 \cup A_2$. $\endgroup$ Oct 15, 2018 at 16:06
  • $\begingroup$ @BenBray Why does $Y_1 = Y_2$ a.s. on $A_i$ (for either $i$)? Why couldn't it be that $\int_{A_1} ( Y-Y') = \int_{A_2}(Y - Y') \neq 0$? $\endgroup$
    – Tom
    Oct 19, 2018 at 13:09

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