I am trying to show that for a probability space $(\Omega,\mathcal{F},\mathbb{P})$, if $\mathcal{G} \subset \mathcal{F}$ is a sigma-field, then
$$ \mathbb{P}(1_A >0) = \mathbb{P}(\mathbb{E}(1_A|\mathcal{G})>0) $$ for any $A\in\mathcal{F}$, where $\mathbb{E}(1_A|\mathcal{G})$ is the conditional expectation. However, I am stuck. Can anyone provide hints or solution?
The equality you are trying to prove does not seem to be true: suppose that $\mathcal G$ is generated by a partition $(B_i)_{i=1}^n$, each of them having positive probability. Then $$ \mathbb E\left[\mathbf{1}_A\mid\mathcal G\right]=\sum_{i=1}^n\mathbf{1}_{B_i}\frac{\mathbb P(A\cap B_i)}{\mathbb P(B_i)} $$ hence denoting $I=I(A)=\{i\in \{1,\dots,n\},\mathbb P(A\cap B_i)\neq 0\}$, we have $$ \left\{\mathbb E\left[\mathbf{1}_A\mid\mathcal G\right]>0\right\} =\bigcup_{i\in I}B_i$$ hence $$ \mathbb P\left(\mathbb E\left[\mathbf{1}_A\mid\mathcal G\right]>0\right)=\sum_{i\in I}\mathbb P(B_i). $$ If $$\mathbb{P}(1_A >0) = \mathbb{P}(\mathbb{E}(1_A|\mathcal{G})>0),$$ we should have that $\mathbb P(A)\in\left\{ \sum_{i=1}^n a_i\mathbb P(B_i), a_i\in\{0,1\}\right\}$ and the latter set has cardinality smaller than $2^n$.
Therefore, a counter-example can be given by taking $\Omega=[0,1]$ endowed with the Borel $\sigma$-algebra and the Lebesgue measure and $B_1=[0,1/2[$, $B_2=[1/2,1]$ and $A=[0,1/3]$.