Almost sure convergence for expected value of indicator function Let $Z$ a random variable which is $\mathscr{G}$-measurable.
If $\mathcal{C}$ is a class of events, closed under intersection which generates the $\sigma$-field $\mathscr{G}$ (i.e $\mathscr{G}=\sigma(\mathcal{C})$) and $\mathbb{E}[{Z \mathbf{1}_B}]=\mathbb{E}[Z]=0$ for every $B \in \mathcal{C} $
and  $\mathbb{E}[{Z \mathbf{1}_B}]=0$ for every $B \in \mathcal{G}$ then how show that :
$Z = 0 $ almost sure.
Proposal
If $Z\geq 0$, then we may argue that
$$
\mathscr{G}\ni \mathcal{C}\mapsto \mathbb{E}[Z \mathbf{1}_{B}] 
$$
are measures that agree on the $\pi$-system $\mathscr{G}$ and thus also on the sigma-algebra $\sigma(\mathcal{C})=\mathscr{G}$.
Now for a general $Z\in L^1(\Omega,\mathcal{C},P)$ we have
$$
\mathscr{G}\ni \mathcal{C} \mapsto \mathrm{E}[Z^{+}|B]+\mathrm{E}[Z^-| B]
$$
are measures that agree on the $\pi-\lambda$-system.
Since $Z$ is $\mathcal G$ measurable an $EZ1_B=0$ for all $B \in \mathcal G$ we have $EZ1_{Z&gt;0}=0$ and $EZ1_{Z&lt;0}=0$. But this implies that $P(Z&gt;0)=0$ and $P(Z&lt;0)=0$ so $Z=0$ almost surely.
Since $Z$ is $\mathcal G$ measurable an $EZ1_B=0$ for all $B \in \mathcal G$ we have $EZ1_{Z&gt;0}=0$ and $EZ1_{Z&lt;0}=0$. But this implies that $P(Z&gt;0)=0$ and $P(Z&lt;0)=0$ so $Z=0$ almost surely.