I'm working through a set of exercises, and am missing a step in the solution presented by the author. The question asks for the proof:
Let $(\Omega, \mathcal{A},\mathbb{P} )$ be a probability space and $\mathcal{G}$ such that $\mathcal{G} \subseteq \mathcal{A}$. Let $A \geq 0$ be a random variable in $(\Omega, \mathcal{A}, \mathbb{P})$. Prove $A > 0 \Rightarrow E \big[ A \big| \mathcal{G} \big] > 0$ almost surely.
The solution: First, note that $$ X = \Big\{ E\big[ A \big| \mathcal{G} \big] \leq 0\Big\} \in \mathcal{G}. $$ Then by definition, $ 0 < E\big[ A \mathbb{1}_{X} \big] = E \big[E \big[ A \big| \mathcal{G} \big]\mathbb{1}_{X} \big] < 0 $, so $ E\big[ A \mathbb{1}_{X} \big] = 0.$ Since $A>0$, $\mathbb{P}(X)=0$.
Why does this last step imply the result?
That an event happens almost surely means that the probability of the event is 1. In your case this translates to $$ E[A\mid \mathcal G] > 0 \text{ a.s.}\quad \Leftrightarrow \quad P(E[A\mid G] > 0) = 1. $$ The latter is equivalent to $$ P(\underbrace{E[A\mid \mathcal G] \leq 0}_X) = 0, $$ i.e. $$ P(X) = 0 \quad \Leftrightarrow \quad P(E[A\mid\mathcal G]>0) = 1 \quad \Leftrightarrow \quad E[A\mid \mathcal G] > 0 \text{ a.s.} $$