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

1. $X_1-X_2$ and $X_1$ are independent
2. $X_1-X_2$ 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?

-

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.
Could you "spell" out the last few sentences of the proof and explain how $S$ being a singleton implies whatever you're trying to prove. I'm not sure if this is a counterexample or you're proving something? What is your conclusion? – user28877 Nov 9 '13 at 23:41
@user710587 Really? This is amusing... Say, how would you translate the fact the OP is asking to prove, that is, that $X_1-X_2$ is a constant, in terms of $S$ the support of the distribution of $X_1-X_2$? – Did Nov 9 '13 at 23:45