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.
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.