The conditional expectation $\mathbb{E}[\,X\,|\,\mathcal{G}\,]$ of a random variable $X\in\mathcal{L}^{1}$ on some probability space $(\Omega,\mathcal{F},\mathbb{P})$ (where $\mathcal{G}\subset\mathcal{F}$) always exists and is unique up to sets of measure zero.

I am wondering what this last part really means. Here is my understanding: $\mathbb{E}[\,X\,|\,\mathcal{G}\,]$ is unique on all $A\in\mathcal{G}$ with $\mathbb{P}(A)=1$. Is this correct? Or does uniqueness only hold on (at least) one $A\in\mathcal{G}$ with full measure?

On a similar note: the (conditional) dominated convergence theorem states that $$\mathbb{E}[\,X_{n}\,|\,\mathcal{G}\,] \quad\rightarrow\quad \mathbb{E}[\,X\,|\,\mathcal{G}\,]\qquad\mathbb{P}\text{-a.s.}$$ for $X_{n}\rightarrow X$ $\mathbb{P}$-a.s. with $X_{n}\leq Y$, $Y\in\mathcal{L}^1$.

Again: Does this mean, that I can choose some set $A\in\mathcal{G}$ with $\mathbb{P}(A)=1$, such that $$\mathbb{E}[\,X_{n}\,|\,\mathcal{G}\,](\omega) \quad\rightarrow\quad \mathbb{E}[\,X\,|\,\mathcal{G}\,](\omega)$$ for all $\omega\in A$?

Your help is highly appreciated! Thank you.


1 Answer 1


Remember that $\mathbb{E}\left[X\mid\mathcal{G}\right]$ is defined to be a $\mathcal{G}$-measurable random variable such that $$\int_A X d\mathbb{P}=\int_A\mathbb{E}\left[X\mid\mathcal{G}\right]d\mathbb{P}$$ for all $A\in\mathcal{G}$. If one changes the value of $\mathbb{E}\left[X\mid\mathcal{G}\right]$ at a set of measure zero, the value of the integral will not change. So saying that $\mathbb{E}\left[X\mid\mathcal{G}\right]$ is unique up to sets of measure zero means that each two random variables which satisfy the above requirements will only differ on a set of measure zero.

As for your second question -- what you wrote is correct.

  • $\begingroup$ Thank you for the clarification! $\endgroup$
    – Mark
    Jun 2, 2017 at 14:59

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