Why is E[X+Y|X,Y]=X+Y? Intuitively, it seems obvious, but I am struggling to prove it for the case where $X_1, ..., X_n$ are continuous random variables. I am aware that $E[c(X)|X]=c(X)$. So how would one show that $E[c(X_i) |X_1, ..., X_n] = c(X_i)$ and that $E[c(X_i)+a(X_j)|X_1, ..., X_n] = c(X_i) + a(X_j)$ for $i,j=1, ..., n$?

You can compute an expectation by summing over all possibilities
$$E[f(X,Y)|\text{give some condition}] = \sum_{\text{for all cases of $x,y$}} f(x,y) \cdot p(X=x,Y=y)$$
When the condition is $X=a$ and $Y=b$, then the summation only contains a single term and $p(X=x,Y=y) = 1$ when $x=a$ and $y=b$.
