Can we use Fubini's Theorem?

Is there any special technique to deal with the distribution of sum of two random variables where they are not independent?

For example I have concluded that if $X =_p W$ and $Y=_pZ$ ($=_p$ means having same distribution) then these two sum must be equal

$$\ \int_{t}P({X+Y< t<W+Z}) = \int_t P({W+Z< t< X+Y}) .$$

But I don't know how to do it technically. It seems to be true by intuition!

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What do you mean by $=_p$? – martini Nov 2 '12 at 10:23
means having same distributions – peanut Nov 3 '12 at 4:34
It would be interesting to know how you have concluded that .../... these two probabilities must be equal, since they are not always equal. – Did Nov 3 '12 at 10:38
Actually my conclusion was $$\ \int_t P({X+Y< t<W+Z}) = \int_t P({W+Z< t< X+Y}) .$$ and I thought that maybe the integrand are equal ( as a stronger guess!) – peanut Nov 3 '12 at 10:38
Briefly put, the pointwise version fails (as demonstrated by @martini) but the integrated version holds (see my answer). – Did Nov 3 '12 at 10:59

On $\Omega = [0,1]$ with the Lebesgue measure, let $X(\omega) = W(\omega) = Z(\omega)=\omega$ and $Y(\omega) = 1- \omega$. Then all four variables are uniformly $[0,1]$-distributed. We have, that $X+Y$ is constant, and $W+Z$ is uniformly $[0,2]$-distributed. No, for say $t= \frac 12$ we have $P\left(X+Y < \frac 12 < W+Z \right) = P\left(1 < \frac 12 < W+Z \right) = 0$ and $P\left(W+Z < \frac 12 < X+Y \right) = P\left(W+Z < \frac 12 < 1 \right) = P\left[0,\frac 14\right) = \frac 14$
Thanks but my fault! The main phrase was the integral of both sides on $t$ from $-/inf$ to $+/inf$. I have updated the right one – peanut Nov 3 '12 at 10:42
Considering $U=X+Y$ and $V=W+Z$, the LHS is $$\int_{-\infty}^{+\infty}\mathbb P(U\lt t\lt V)\,\mathrm dt=\mathbb E((V-U)^+),$$ and the RHS is $$\int_{-\infty}^{+\infty}\mathbb P(V\lt t\lt U)\,\mathrm dt=\mathbb E((U-V)^+).$$ The LHS and the RHS coincide as soon as $\mathbb E(U)=\mathbb E(V)$ since, for every integrable random variable $R$, $\mathbb E(R)=\mathbb E(R^+)-\mathbb E((-R)^+)$. Finally, $X\stackrel{(d)}{=}W$ and $Y\stackrel{(d)}{=}Z$ hence, if all these random variables are integrable, then $\mathbb E(X)=\mathbb E(W)$ and $\mathbb E(Y)=\mathbb E(Z)$, which implies $\mathbb E(U)=\mathbb E(V)$, hence we are done.