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Consider $X$ as non-decreasing non-negative function. Consider $\mu$ and $\nu$ as two probability measures on $(\mathbb{R},\mathcal{B})$ for which we know $\mu([t, \infty)) \geqslant \nu([t, \infty)) \;\forall t \in \mathbb{R}$. How can I show that $\int Xd\mu \geqslant \int Xd\nu$?

I started from the definition of the integral of non-negative function but was not successful proving it. Then I thought since $X$ is monotone, maybe monotone convergence can help, but after 3 hours, I still don't know how to approach this. I appreciate if you could guide me.

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I think the basic idea is to approximate $X$ with functions $\sum r_{t_i} 1_{[t_i, \infty)}$, where $1_{[t_i, \infty)}$ is the eigenfunction on $[t_i, \infty)$. – ougao Dec 13 '12 at 18:35

1 Answer 1

up vote 1 down vote accepted

Let $F(\omega,t):=\chi_{S}(\omega,t)$, where $S:=\{(\omega,t), X(\omega)\geqslant t\}$. Using Fubini's theorem, we get $$\int_\Bbb R X(\omega)d\mu(\omega)=\int_{\Bbb R}\int_{\Bbb R}\chi_S(\omega,t)dtd\mu(\omega)=\int_{\Bbb R}\mu\{\omega,X(\omega)\geqslant t\}dt.$$ Do the same for $\nu$ and compare.

Note that we just need $X$ to be non-negative.

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