Prove $E(X) = \sum_{j=1}^n E(X|X \in A_j) \mathrm{Pr}(X \in A)$ I'm looking to prove:
Let X denote a real-valued random variable with range X, such that E(|X|) < $\infty$. Let $A_{1}$,...$A_{n}$ denote disjoint subsets of X. Show that
$E(X) = \sum_{j=1}^n E(X|X \in A_j ) \mathrm{Pr}(X \in A)$
However this equation is quite difficult to solve. 

I think there is a typo in your formula. You want to prove
$$E(X)=\sum E(X|A_{j})P(A_{j})$$
instead, here $A_{j}$ is a sequence of events such that $P(\bigcup A_{j})=1$. But I think this follows from the definition of conditional expectation. Maybe checking your textbook will be helpful. 
Update:
As pointed out by Did, there is an ambiguity. I think the classical formula $E(X|Y)$ is only making sense as the expectation of $X$ relative to $Y$, which means this is a function of $Y$. I suspect OP means $A_{j}$ to be a partition of the domain and $E(X|A_{j})$ is the expectation of $X|A_{j}$ on $A_{j}$. But then we have
$$E(X)=\int xf_{X}(x)dx=\bigcup \int_{A_{j}} xf_{X}(x)dx=\sum E(X|A_{j})P(A_{j})$$
The last equation follows because $X|A_{j}$ is the restriction of $f_{X}(x)$ at $A_{j}$ divided by $P(A_{j})$. 
