stochastic dominance
Let $F$ stochastically dominates $G$.
Then, (1)$F(x)\le G(x) \forall x$, and (2)$\int u dF \ge \int u dG$ where $u$ is an increasing function.
Proof from (1) to (2): Let $x\in [a,b],$ and $u$ is right continuous . Then,
$\int_a^b u(x)dF(x) = u(x)F(x)|_a^b-\int_a^b F(x)du(x) = u(b)-\int_a^b F(x^-)du(x)$.
Similarly, $\int_a^b u(x)dG(x)=u(b)-\int_a^b G(x^-)du(x)$.
Then, (1) implies that $\int_a^b G(x^-)du(x)\le\int_a^b F(x^-)du(x)$.
But, I don't understand what it means. How (1) implies that result?
Also, what does $F(x^-)$ means? why does the function have $x^-$ as a variable?