What can be said about $E[1_A\mid\mathcal F]$?

Question

It is known that $E[X\mid 1_A]$ is of particularly nice form. What can be said about the form of $E[1_A\mid\mathcal{F}]$ for "general" $\mathcal{F}$? Is it true that $E[1_A\mid\mathcal{F}]=1_B$ for some $B\in \mathcal{F}$?

Answer 1

If $\Omega$ is a partition which consists of sets of positive measure, say $(S_i)_{i\in \Bbb N}$, with $S_i\in\mathcal F$, then $$E[\chi_A\mid\mathcal F]=\sum_{i\in\Bbb N}\mu(A\cap S_i)\chi_{S_i}\mbox{a.s.},$$ hence $E[\chi_A\mid\mathcal F]$ is not necessarily the characteristic function of a measurable set $B\in\mathcal F$. It's even not necessarily a simple function.

If $A$ is independent of $\mathcal F$, then $E[\chi_A\mid \mathcal F]=\mu(A)$ a.e. (in particular, not a characteristic function).

Answer 2

For every $A$ and $\mathcal F$, $E[1_A\mid\mathcal F]$ is $[0,1]$ valued. Conversely, every $[0,1]$ valued random variable can be realized as $E[1_A\mid\mathcal F]$ for some $A$ and $\mathcal F$.

Direct part: $0\leqslant 1_A\leqslant1$ almost surely hence $0=E[0\mid\mathcal F]\leqslant E[1_A\mid\mathcal F]\leqslant E[1\mid\mathcal F]=1$ almost surely.

Converse part: fix some $[0,1]$ valued random variable $X$ with CDF $F:x\mapsto P(X\leqslant x)$. Consider $\Omega=[0,1]^2$ endowed with its Borel sigma-algebra, $\mathcal F$ the sigma-algebra generated by the second projection and $A=\{(x,y)\in\Omega\mid y\geqslant F(x)\}$. Then $Z=E[1_A\mid \mathcal F]$ is such that $Z(x,y)$ depends on $y$ only, and one easily checks that $Z(x,y)=\mathrm{Leb}\{z\mid y\geqslant F(z)\}$ for almost every $x$. Hence, the properties $Z(x,y)\leqslant z$ and $y\leqslant F(z)$ are equivalent. In particular, $P[Z\leqslant z]=F(z)$.