When does $\mathbb E_\mathbb P[X]=0$ imply $\mathbb E_\mathbb P[X\mid\mathcal E]= 0$ $\mathbb P$-a.s.

Question

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, $\mathcal E\subseteq\mathcal F$ a sub-$\sigma$-algebra, $X:\Omega\rightarrow\mathbb R^d$ a $\mathcal F$-measurable map. My question is, when does $\mathbb E_\mathbb P[X]=0$ imply that $\mathbb E_\mathbb P[X\mid\mathcal E]= 0$ $\mathbb P$-a.s. ? Clearly, if $\mathcal E$ is trivial, then this implication holds. But what are other necessary and sufficient conditions on $\mathbb P$ and/or $\mathcal E$ in order that this implication holds? For example, does the implication hold if we assume that $\mathbb P$ is of the form $\mathbb P=\sum_{i=1}^Na_i\delta_{\omega_i}$ for $a_i\in (0,1]$, $N\in\mathbb N\cup\{\infty\}$?

Answer 1

Assume that $\mathbb P$ and $\cal E$ are such that the implication holds. If $X$ is $\cal E$-measurable, define $X':=X-\mathbb E(X)$. Then we should ahve $$\mathbb E_{\mathbb P}[X'\mid \mathcal E]=0,$$ hence $X$ is $\mathbb P$-almost surely constant. Therefore, considering characteristic functions, $\mathbb P(E)\in\{0,1\}$ for each $E\in\mathcal E$.