How do I show that $\displaystyle\sum_{x\in\text{Im}(X)}xP[X=x\mid A] = \dfrac{E[X\cdot1_A]}{P[A]}$, for $A \subseteq \Omega$ a sub-sigma algebra and $X$ a discrete random variable from $\mathcal{F} \to \mathbb{R}$?
The hint was that we should start by assuming that $X$ takes finite many values, and then extend the result to every $X\ge0$. Also, we can write $X(\omega) = \displaystyle\sum_{x\in\text{Im}(X)}x1_{X=x}(\omega)$.
I attempted to expand both sides to get $$\displaystyle\sum_{x \in \text{Im}(X)} x\mathbb{P}[X = x \mid A] = \displaystyle\sum_{x \in \text{Im}(X)} x\dfrac{\mathbb{P}[X = x \cap A]}{\mathbb{P}[A]}$$ and $$\dfrac{E[X\cdot1_A]}{\mathbb{P}[A]} = \displaystyle\sum_{x \in \text{Im}(X)} x\dfrac{\mathbb{P}[X\cdot1_A = x]}{\mathbb{P}[A]},$$ then trying to show that for every $x \in \text{Im}(X)$, $\mathbb{P}[X = x \cap A] = \mathbb{P}[X\cdot1_A = x]$. But this is where I got stuck, since $X\ge0$ and not every $\omega$ s.t. $X\cdot1_A(\omega) = x$ is in $A$ if $x = 0$. Also, I'm not entirely sure why is it that we should start by assuming $X$ takes only finite values.
Any help is appreciated!
\begin{gather*} E[X\cdot 1_A]=E\left[\sum x1_{X=x}\cdot 1_A\right]=\sum xE[1_{X=x}\cdot 1_A]=\sum xE\bigl[E[1_{X=x}|A]\cdot1_A\bigr]=\\ =\sum xE[1_{X=x}|A]E[1_A]=\sum xP[X=x|A]P(A) \end{gather*} Another way: \begin{gather*} \sum xP[X=x|A].P(A)=\sum xP(X=x\cap A)=\sum xE[1_{X=x}\cdot 1_A]=E\left[\sum x1_{X=x}\cdot 1_A\right]=E[X\cdot 1_A] \end{gather*} In general: $$E[X|A]=\int xdP_{X/A}=\frac{E[X\cdot 1_A]}{P(A)}$$