Independence implies constancy Given a probability space $(\Omega,\mathcal{F},P)$ on which we define two random variables $X_1$ and $X_2$.
From the following two independence conditions


*

*$X_1-X_2$ and $X_1$ are independent 

*$X_1-X_2$ and $X_2$ are independent 


How could we deduce that $X_1 - X_2$ is constant almost surely?
I only managed to deal with the case when both $X_1$ and $X_2$ are square integrable:
\begin{align}
E[(X_1 -X_2)^2] &= E[(X_1 -X_2)X_1] - E[(X_1 -X_2)X_2] \\ &= E[X_1-X_2]E[X_1] - E[X_1 -X_2]E[X_2] \\&= (E[X_1 -X_2])^2
\end{align}
which means the variance of $X_1 -X_2$ is zero, so $X_1 -X_2$ is a constant almost surely.
But in general, without assumption of integrability on $X_1$ and $X_2$, how could we prove the conclusion? Or is the conclusion still true?
 A: Let $\varphi$ denote the characteristic function of $X_1-X_2$, that is, $\varphi(t)=E[\mathrm e^{\mathrm it(X_1-X_2)}]$ for every real number $t$. The independence of $X_1-X_2$ and $X_2$ yields
$$E[\mathrm e^{\mathrm itX_1}]=E[\mathrm e^{\mathrm it(X_1-X_2+X_2)}]=\varphi(t)\cdot E[\mathrm e^{\mathrm itX_2}].
$$
Likewise, the independence of $X_1-X_2$ and $X_1$ yields
$$E[\mathrm e^{\mathrm itX_2}]=\varphi(-t)\cdot E[\mathrm e^{\mathrm itX_1}].
$$
Thus,
$$
E[\mathrm e^{\mathrm itX_1}]=\varphi(t)\cdot \varphi(-t)\cdot E[\mathrm e^{\mathrm itX_1}]=|\varphi(t)|^2\cdot E[\mathrm e^{\mathrm itX_1}].
$$
For every $t$ such that $E[\mathrm e^{\mathrm itX_1}]\ne0$, in particular, for every $t$ in a neighborhood of $0$, one sees that $$|\varphi(t)|=1.$$ 
This implies that, for every $|t|$ small enough, the support $S$ of the distribution of $X_1-X_2$ is included in $x_t+2\pi\mathbb Z/t$ for some $x_t$. Since this holds for every $t$ when $t\to0$, $S$ intersects every finite interval on at most one point. Thus, $S$ is a singleton.
