# How is this use of Fubini's theorem justified?

In the paper "A Comparison Theorem on Moment Inequalities Between Negatively Associated and Independent Random Variables" (Qi-Man Shao, 2000) the following theorem is proved (paraphrased):

Let $$X_1, \dots, X_n$$ be negatively associated random variables, and let $$X_1^*, \dots, X_n^*$$ be corresponding independent random variables such that each $$X_i$$ and $$X_i^*$$ have the same distribution. Then $$\mathbf{E}\left[f\left(\sum_i X_i\right)\right] \le \mathbf{E}\left[f\left(\sum_i X_i^*\right)\right]$$ for any convex function $$f : \mathbb{R} \to \mathbb{R}$$, whenever the expectation on the right hand side exists.

They give the following proof when $$n = 2$$:

("H2" is the Fundamental Theorem of Calculus for convex functions)

I am having trouble understanding how they ensure that Fubini's theorem is applicable here. I was able to show that if we make the stronger assumption that both of the expectations in the theorem statement are finite, then Fubini's theorem applies because the integrand is absolutely integrable:

\begin{align*} &\frac12\mathbf{E}\left[\int_{-\infty}^\infty |(f'(Y_1 + t) - f'(X_1 + t))(I_{t < Y_2} - I_{t < X_2})|\,dt \right] \\ &= \frac12\mathbf{E}\left[\left|\int_{-\infty}^\infty (f'(Y_1 + t) - f'(X_1 + t))(I_{t < Y_2} - I_{t < X_2})\,dt \right|\right] \\ &= \frac12\mathbf{E}[|f(X_1 + X_2) + f(Y_1 + Y_2) - f(X_1 + Y_2) - f(Y_1 + X_2)] \\ &\le \mathbf{E}[|f(X_1 + X_2)|] + \mathbf{E}[|f(X_1^* + X_2^*)|] < \infty & \text{by assumption} \end{align*} where the first equality is because both $$x \mapsto f'(x + t)$$ and $$x \mapsto I_{t < x}$$ are increasing, so for fixed $$X_1, Y_1, X_2, Y_2$$, the integrand has the same sign for all $$t$$.

But I cannot figure out how to make it work using only the assumption that the RHS expectation exists. How can this be done?

• Clamp the $$X_i$$ and $$X_i^*$$ to the range $$[-n, n]$$ so that all the expectations are finite, and use the original argument by Fubini's theorem to prove the statement in this case.
• Extend the statement to only require the precondition that both expectations exist, by using monotone convergence to take the limit as $$n \to \infty$$ in the above. When $$f$$ is bounded from below this is easy, but otherwise we have to argue that one tail is decreasing and the other is increasing, and make separate monotone convergence arguments for the two sides.
• Show that if $$X, Y$$ are random variables such that $$\mathbf{E}[f(X)] \le \mathbf{E}[f(Y)]$$ for convex $$f$$ whenever both expectations exist, then in fact $$\mathbf{E}[f(X)]$$ exists whenever $$\mathbf{E}[f(Y)]$$ does. This recovers the original theorem statement. Here we consider convex functions $$f$$ which aren't bounded from below, and use the fact that asymptotically $$f$$ decreases towards $$-\infty$$ at a linear rate.