Is the sum of two variables independent of a third variable, if they are so on their own? Given 3 random variables $X_1$, $X_2$ and $Y$. $Y$ and $X_1$ are independent. $Y$ and $X_2$ are independent. Intuitively I would assume that $Y$ and $X_1+X_2$ are independent. Is this the case, and how can I prove it formally?
 A: EDIT: As pointed out by other users, this answer is not correct because it makes the assumption that $Y$ is independent of $(X_1,X_2)$
Note that $X_1 + X_2$ is a function of $Z = (X_1,X_2)$ because if you take 
$$f(x,y) = x+y$$
you get $X_1 + X_2 = f(Z)$.
It is a well known theorem of probability that if $R_1$ and $R_2$ are independent random variables and $f_1$ and $f_2$ are measurable functions then $f_1(R_1)$ is independent of $f_2(R_2)$ (Theorem 10.4 of "Probability: A Graduate Course" 2nd ed. by Allan Gut).
Since $f$ is measurable and Y is independent of $Z$ we know that $Y$ is also independent of $f(Z) = X_1 + X_2$. Note that we took $f_1$ as the identity function and $f_2 = f$.
A: (To complete this thread, I am elevating a comment by user233740 into an answer.)
The statement is not true.
The possibility that $X_1+X_2$ might not be independent of $Y$ is strongly reminiscent of the familiar textbook problem concerning trivariate random variables $(X_1,X_2,Y)$ that are pairwise independent but not independent.  With that thought in mind, let's consider the simplest such example, that of selecting one of the rows of this matrix uniformly at random:
$$\pmatrix{0&0&0\\1&1&0\\1&0&1\\0&1&1}.$$
You can see that any two columns determine independent Bernoulli$(1/2)$ variables, but the three are not independent because the third can be determined from the other two.
Let us then pick two of these columns, denoting them $X_1$ and $X_2,$ and let $Y$ be the third.  Observe that when $Y=0,$ $X_1+X_2$ is either $0$ or $2$ (with equal probability), but when $Y=1,$ $X_1+X_2=1.$  Thus the conditional probability function $$\Pr(X_1+X_2\mid Y)$$ is not constant, demonstrating $X_1+X_2$ and $Y$ are not independent.
