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 $i,j=1, ..., n$?

In general if $\mathcal A$ is a $\sigma$-algebra on the probability space and a random variable $Z$ is measurable wrt $\mathcal A$ then:$$\mathbb E[Z\mid\mathcal A]=Z$$
Here $X$ and $Y$ are both measurable wrt $\sigma(X,Y)$ and consequently $X+Y$ is measurable wrt $\sigma(X,Y)$.
So in this situation:$$\mathbb E[X+Y\mid X,Y]:=\mathbb E[X+Y\mid\sigma(X,Y)]=X+Y$$
This can be extended to:$$\mathbb E[f(X,Y)\mid X,Y]=f(X,Y)$$for any Borel-measurable function $f:\mathbb R^2\to\mathbb R$.
