TITLE: Proof: $Y$ stochastically dominates $X$ implies $E[\phi(Y)]\geq E[\phi(X)]$ for increasing $\phi$
QUESTION [6 upvotes]: Suppose $X$ and $Y$ are real random variables with CDF $F$ and $G$ such that $F(x)\geq G(x)$ (i.e. $Y$ exhibits (first-order) stochastic dominance over $X$). Then, for all increasing function $\phi$, we have $E[\phi(Y)]\geq E[\phi(X)]$. I would like to find a rigorous proof of this. Can you please provide one or point me to a reference?


For the special case in which both $F$ and $G$ are continuous, one can consider the integral
$$
\int \phi(y)dG(y)
$$
and employ the substitution $y\to\varphi(x)$ where $\varphi(x)\equiv G^{-1}[F(x)]$ to infer that
$$
E[\phi(Y)]=\int\phi(y)dG(y){\color{red}=}\int\phi[\varphi(x)]dF(x)\geq\int\phi(x)dF(x)=E[\phi(X)].
$$
The inequality above is because
$$
F(x)\geq G(x)\implies\varphi(x)=G^{-1}[F(x)]\geq G^{-1}[G(x)]{\color{blue}=}x;\quad \phi\text{ is increasing.}
$$
I've made some attempt for when $F$ and $G$ are general by considering $\varphi(x)=G^*[F(x)]$ where $G^*$ now is the (left-continuous) quantile function of $Y$. I've run into at least 2 problems: (i) I don't think the equality ${\color{red}=}$ above continues to hold; (ii) $G^*[G(x)]\leq x$ so ${\color{blue}=}$ above is turned into $\leq$ which breaks the inequality chain.
A rigorous proof can also follow from a different direction. But this different direction relies on a construction I don't know how to do.

REPLY [2 votes]: Here is an easier proof.
Since increasing functions are differentiable almost surely, we may consider all $\phi$ that is both differentiable and increasing without loss of generality.
Define
$$
H(x)=G(x)-F(x)\leq 0,\quad H(0)=0,\quad \lim_{x\rightarrow \infty}H(x)=0.
$$ 
Apply the Integration by Parts for the generalized distribution function
$$
\int \phi(x)dH(x)=[\phi(x)H(x)]^\infty_{0}-\int \phi'(x)H(x-)dx
$$
where
$$
H(x-)=\lim_{t\uparrow x}H(t)\leq 0.
$$
The first term vanishes by construction. The second term is evidently positive. It follows that
$$
\int \phi(x)dG(x)-\int \phi(x)dF(x)=\int \phi(x)dH(x)\geq 0.
$$
This completes the proof.