If $\mathbb E[X\boldsymbol 1 _G]=0$ for all $G\in \mathcal G$, does $X=0$?

Question

Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $\mathcal G$ a subfield of $\mathcal F$. I have that $\mathbb E[X\boldsymbol 1_G] = 0$ for all $G\in \mathcal G$. Do we have that $X=0$ ? I proved that if $X\geq 0$ a.s. then $X=0$ a.s. but if $X$ is just measurable, then I have that $X=X^+-X^-$ and thus $$\int_{G}X^+=\int_GX^-,$$ but how can I conclude ?

Answer 1

This holds true, if $X$ is not only $\def\F{\mathcal F}\F$-, but $\def\G{\mathcal G}\G$-measurable. For general $\F$-measurable functions it is wrong, as shown in the other answer. Hence suppose, that $X$ is $\G$-measurable. Let $n\in\mathbf N$, and $G := \{X > \frac 1n\}\in \G$, as $X$ is $\G$-measurable. Therefore $$ 0 = \def\E{\mathbf E}\E[X1_G] = \int_{X > \frac 1n} X\, d\def\P{\mathbf P}\P \ge \frac 1nP\left(X > \frac 1n\right) \iff \P\left(X > \frac 1n\right) = 0 $$ Hence $$ \P(X > 0) = \lim_n \P\left(X > \frac 1n\right) = 0 $$ With $G = \{X < \frac {-1}n\}$, we obtain, along the same lines, that $$ \P(X < 0) = \lim_n \P\left(X < -\frac 1n\right) = 0$$ Therefore $\P(X\ne 0)= 0$, that is $X = 0$ $\P$-a. s.

Answer 2

If $\mathcal{G} = \{\emptyset,\Omega\}$, then the only information you have is that $\mathbb{E} X = 0$. In general, this is not enough to conclude that $X = 0$. (However, if $X \geq 0$ then $\mathbb{E}X=0$ does imply that $X=0$ almost surely.)