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I saw an amazing (I think) tool in one of the exercises of Durrett (page 225) which I am going to use it as a lemma to solve another (amazing) problem. The Durrett's is:

Ex4.1.4. Suppose $X\geq 0$ and $EX= \infty$. Show that there is a unique $F$-measurable $Y$ with $0\leq Y\leq\infty$ so that

$\int_A X \,dP = \int_A Y \,dP $ for all $A\in F$.

With its hint this problem becomes easy. But now my problem is if we have $Z_1, Z_2, Z_3$ and $Z_4$, four random variables where none of them have bounded expectation, and we know that $E[Z_1 + Z_2] = 0$ and $E[Z_3 + Z_4] < \infty$ and that $Z_1 \stackrel{d}{=} Z_3$ and $Z_2 \stackrel{d}{=} Z_4$. Can we now conclude that $E[Z_3 + Z_4]= 0$?

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Something appears to have gone missing from your problem statement. What edition of Durrett are you using and can you cite a page number? – cardinal Nov 29 '12 at 19:37
Your "amazing tool" is better known as "conditional expectation". But I'm not sure I see how it will be helpful in solving your problem. – Nate Eldredge Nov 29 '12 at 19:40
the point is that this conditional expectation is for a random variable with infinite expectation – Heinse Nov 29 '12 at 19:43
up vote 2 down vote accepted

Consider the truncation function $$t_n(x) = \cases{-n & if $x < -n$\cr x & if $-n \le x \le n$\cr n & if $x > n$\cr}$$

Note that $|t_n(x) + t_n(y)| \le |x + y|$ for all real $x,y$.
For any random variables $X$ and $Y$ such that $E[X+Y]$ exists, by the Dominated Convergence Theorem we have $E[t_n(X) + t_n(Y)] \to E[X+Y]$ as $n \to \infty$. Now in your case $$E[t_n(Z_1) + t_n(Z_2)] = E[t_n(Z_1)] + E[t_n(Z_2)] = E[t_n(Z_3)] + E[t_n(Z_4)] = E[t_n(Z_3) + t_n(Z_4)]$$ so taking the limit as $n \to \infty$, if $E[Z_1 + Z_2]$ and $E[Z_3 + Z_4]$ exist we must have $E[Z_1 + Z_2] = E[Z_3 + Z_4]$.

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Which integrable random variable dominates every $|t_n(X)+t_n(Y)|$, to apply DCT? – Did Nov 29 '12 at 19:57
$|t_n(X) + t_n(Y)| \le |X+Y|$ – Robert Israel Nov 29 '12 at 20:22
Right, each $t_n$ is odd and $1$-Lipschitz. Nice. – Did Nov 29 '12 at 21:28

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