$\exists f:\mathbb{R}\to\mathbb{R}$ such that $\frac{f(X_1)}{X_1+X_2}$ is independent of $X_1$? Let $X_1, X_2$ be independent real random variables. Does there exist a non-zero measurable function $f:\mathbb{R}\to\mathbb{R}$ such that $$\frac{f(X_1)}{X_1+X_2}\text{ is independent of }X_1?$$
I am especially interested in the case when $X_1, X_2$ are continuous, but we do not assume anything else about them.
My ideas are the following: if $\frac{f(X_1)}{X_1+X_2 }$ does not depend on $X_1$, then $\frac{f(X_1)}{X_1+X_2}=g(X_2)$. Rewriting this we get $X_1+X_2 = f(X_1)g^{-1}(X_2)$. But on the left-hand side, we have a sum, and on the right side a product. Hence, that's a contradiction.  However, these steps are not rigorous and I am not really sure if I can make them. Especially, the existence of $g$ probably does not hold.
 A: If we are interested in just existence, there is such example. Let $\xi = \frac{f(X_1)}{X_1 + X_2}$.
Assume $X_1$ takes values $x_1$ and $x_2$ with probabilities $\frac{1}{2}$ each, and $X_2$ takes $y_1$ and $y_2$.
Then, if $X_1 = x_1$, $\xi$ takes values $\frac{f(x_1)}{x_1 + y_1}$ and $\frac{f(x_1)}{x_1 + y_2}$ with probability $\frac{1}{2}$, and if $X_1 = x_2$ - values $\frac{f(x_2)}{x_2 + y_1}$ and $\frac{f(x_2)}{x_2 + y_2}$.
As we want $\xi$ to be independent of $X_1$, the sets of values should be the same.
Assume that, like in Yuval's example in comments, different values of $X_1$ transposes values of $\xi$, so we have
$$\begin{cases}
\frac{f(x_1)}{x_1 + y_1} = \frac{f(x_2)}{x_2 + y_2}\\
\frac{f(x_1)}{x_1 + y_2} = \frac{f(x_2)}{x_2 + y_1}
\end{cases}$$
This is equivalent to
$$\begin{cases}
x_1 + y_1 + x_2 + y_2 = 0\\
f(x_1) = -f(x_2)
\end{cases}$$
and none of denominators above being equal to $0$.
So, we can take, for example, $x_1 = 1$, $x_2 = -1$, $f(x) = x$, $y_1 = 2$, $y_2 = -2$, and get
$$P\left(\xi = \frac{1}{3}\right) = P\left(\xi = \frac{1}{3} | X_1 = 1\right) = \frac{1}{2}$$
$$P\left(\xi = -1\right) = P\left(\xi = -1 | X_1 = 1\right) = \frac{1}{2}$$
Thus $\xi$ is independent of $X_1$.
