Conditional expectation given event and almost sure equality

Question

Let $(\Omega, \mathcal{A}, P)$ be a probability space and $X$ a numerical fct that is either non-negative and measurable or integrable. Let $H$ be an event then we can define for $P(H)>0$ $$ \mathbb{E}[X|H]=\frac{1}{P(H)}\int_H X dP$$ and for $P(H)=0$ we can just set $\mathbb{E}[X|H]=c$, where $c\in\mathbb{R}$.

Now this can be related to the conditional expectation of the sigmal algebra generated by $H$, $\sigma(H)=\{\emptyset, H, H^c\Omega\}$. It holds for 0ドル<P(H)<1$: $$\mathbb{E}[X|\sigma(H)](\omega)=\mathbb{E}[X|H]\chi_{H}(\omega)+\mathbb{E}[X|H^c]\chi_{H^c}(\omega).$$

Now I wonder if this equation holds for every $\omega\in\Omega$ or just for almost all $\omega\in\Omega$ (the latter being claimed in the answers here: Conditional expectation given an event is equivalent to conditional expectation given the sigma algebra generated by the event

For the equality to only hold almost everywhere there would have to be a nullset $N\in\sigma(H)$, such that the equaility holds for all $\omega\in N^c$. Now the empty set is the only null set in that sigma algebra hence I deduce that the above equality holds $\textit{for all } \omega\in\Omega$. Is that reasoning correct?

Now for $P(H)=1$ we have $$\mathbb{E}[X|\sigma(H)](\omega)=\mathbb{E}[X|H]\chi_{H}(\omega). $$ Here since $P(H)=1$ it follows that $H^c$ is a nullset that may be non-empty. Now I dont know if the exepctional nullset on which the equality does not hold can be $H^c$? I.e., can I make a statement here about whether or not this holds for all $\omega\in\Omega$ or just for almost every $\omega\in\Omega$?

Now on $H$ we have that $$\mathbb{E}[X|\sigma(H)](\omega)=\mathbb{E}[X|H].$$ In the answers in the above link they claim that it holds only for almost every $\omega\in H$. But I would argue that for the case $P(H)>0$ this will always hold for all $\omega in H$, since the expectional nullset cannot be a subset of $H$. Is this reasoning correct?

Finally, in the book "Probability theory" by Bauer, he deals with the example where the sigma algebra $\mathcal{C}$ is generated by a finite or countable partition $(B_n)_n$ of $\Omega$ and he states that $$ \mathbb{E}[X|\mathcal{C}]=\sum_{n\in\mathbb{N}}\mathbb{E}[X|B_n]\chi_{B_n}$$ holds $P$-almost surely for $P(B_n)>0$ for all $n\in\mathbb{N}$. Can this be correct? I.e., if all the set have positive measure, can there be a nullset in the generated sigma algebra other than the emptyset?

• userא0: the conditional expectation given a $\sigma$-algebra is unique a.s (by the Radon Nikodym theorem) so the equality has be up to an event of probability 0ドル$.
• guest1: @userא0 I know but the null set has to be in the sub sigma algebra that we condition on. Hence, if the only null set in that sigma algebra is the empty set the equality should hold for every $\omega\in\Omega$
• Shaikh Ammar: It was only claimed that for continuous conditioning variables, the expectation is defined ae. For the conditioning on non-null sets, the definition must be unique otherwise one of the axioms is violated. For null sets, you can define it as whatever you like.
• Shaikh Ammar: Sorry, I thought you were referring to my recent answer on your question (lots of similar questions with similar titles). It's true that the answer you linked is incorrect.

Answer 1

I think the different answers to the question that you linked to throw valuable light on your problem.

The accepted answer states that the problem in the book requires a proof of exactly your conclusion that the equality holds for all $\omega\in H$ and not just almost all $\omega$.

The top answer says that it is not necessarily true that $E(X,円\vert,円 \mathcal{F})(\omega) = E(X,円\vert,円 H)$ for all $\omega \in H$. But if you read carefully, you will observe that

• The conditional expectation has been computed with respect to $\mathcal{F}$ and not $\mathcal{H}$.
• There are two different definitions of $\mathcal{F}$ within the same answer. First that $\mathcal{F}\subseteq \mathcal{H}$ and second that $\mathcal{F} = \mathcal{H}$. (In some situations, considering a $\mathcal{F}$ obtained by augmenting $\mathcal{H}$ with some null sets might be useful and so there might be some merit in considering the first definition.)

Under the first definition, $\mathcal{F}\subseteq \mathcal{H}$, the conditional expectation is not unique and is defined only a.s. I think the author is unconsciously carrying through this non uniqueness to the second case where $\mathcal{F} = \mathcal{H}$ has no non empty null sets and the conditional expectation is unique.

I think in general, the idea that the conditional expectation is defined only a.s. becomes ingrained among probabilists, and this idea permeates their thinking even when there are no non empty null sets in the sigma field.

The case where $P(H)=1$

In this case, the equality holds only for almost all $\omega$ because we can set $\mathbb{E}[X|\sigma(H)]$ to any arbitrary value on $H^c$. For example, consider the uniform distribution on the closed unit interval so that $\Omega=[0,1]$, $\mathcal A$ consists of all the Lebesgue measurable sets and $P$ is Lebesgue measure. Let the random variable $X$ be defined as:

$$X(\omega) = -1, \omega < 0.5 \\ = +1, \omega > 0.5 \\ = 0, \omega = 0.5 \\ $$

Let $H^c$ be the null set $\omega=0.5$. Then conditioning on the event $H$ gives the conditional expectation as 0ドル$ (same as the unconditional expectation). But conditioning on the sigma field $\sigma(H)$, we have many variants of the conditional expectation. If we define $$f_a(\omega)= 0, \omega \ne 0.5\\ = a, \omega = 0.5$$ then for each real number $a$, we find that $f_a$ is also a valid conditional expectation with respect to $\sigma(H)$. Out of these, $f_0$ is a conditional expectation with respect to the sigma field $\sigma(H)$ that agrees with the conditional expectation with respect to the event $H$ for all $\omega$. But, for $a\ne 0$, we find that $f_a$ equals $f_0$ for almost all $\omega$, but not for all $\omega$.

The conclusion is that if $P(H)=1$, the equality holds only for almost all $\omega$, but if 0ドル<P(H)<1$, the equality holds for all $\omega$.

• guest1: R Varna thank you fo ryour answer! Could you comment on the questions I have raised in my question: I.e. is my reasoning that for 0ドル<P(H)<1$: $\mathbb{E}[X|\sigma(H)](\omega)=\mathbb{E}[X|H]\chi_{H}(\omega)+\mathbb{E}[X|H^c]\chi_{H^c}(\omega).$ should hold for all $\omega\in\Omega$ since the null set has to be in the subsigma algebra $\sigma(H)$ and can therefore only be the empty set is correct?
• Jayanth R Varma: Yes. The conditional expectation has to be measurable with respect to $\sigma(H)$ and there is no non empty null set in this sigma field on which the conditional expectation can take a different value from the right hand side.
• guest1: and can you say something regarding my question about the situation : for $P(H)=1$ we have $\mathbb{E}[X|\sigma(H)](\omega)=\mathbb{E}[X|H]\chi_{H}(\omega). $ Does this hold everywhere or only almost everywhere (see my detailed concern in my question)
• Jayanth R Varma: If $P(H)=1,ドル then we can set $\mathbb{E}[X|\sigma(H)]$ to any arbitrary finite value on $H^c$ without affecting the defining requirement for conditional expectation with respect to a sigma field.
• Jayanth R Varma: Yes. That is correct because $\sigma(H)$ does not contain any non empty null subset of $H$.