Prove that if $\mathbb{E}[g(Y)]\leq \mathbb{E}[g(X)]$ for every nondecreasing function $g$ then $F_X\le F_Y$

Question

Let $X,Y$ random variables. Every coninuous and not-decreasing monotone function $g:\mathbb{R}\to[0,1]$ fulfills $\mathbb{E}[g(Y)]\leq \mathbb{E}[g(X)]$. Prove that $F_X(t)\leq F_Y(t)$.

I don't have any initial idea to solve it.

Answer 1

Hint: For a fixed $t$ and $\epsilon>0$, define $g_\epsilon(x)$ as $$g_\epsilon(x)=\begin{cases}0,\quad x\le t\\\frac{x-t}{\epsilon},\quad t<x\le t+\epsilon\\1,\quad t+\epsilon<x\end{cases}. $$ We can see that $g_\epsilon \uparrow 1_{(t,\infty)}(x)$ as $\epsilon\to 0$. Now, by the assumption it holds that $$ E[g_\epsilon(Y)]\leq E[g_\epsilon(X)],\quad\forall \epsilon>0. $$ From this fact, deduce that $$ P(t<Y)=E[1_{\{t<Y\}}]\le E[1_{\{t<X\}}]=P(t<X) $$ for all $t\in\mathbb{R}$. (e.g. by monotone convergence theorem)

Answer 2

HINT:

Use Integration by Parts of the Riemann-Stieltjes Integrals

$$ \int_{-\infty}^\infty g(t)\,dF_X(t)$$

and

$$ \int_{-\infty}^\infty g(t)\,dF_Y(t)$$

Then use the fact that $g$ is increasing and $F_X$ and $F_Y$ are non-negative.