I need to show this proof
$$\int_0^1\int_0^xF(z)dzdx\le \int_0^1\int_0^xG(z)dzdx$$ if and only if $F$ has Lower variance than $G$ for the same mean of $F$ and $G$ and with support $[0,1]$.
Also I know that by the second order stochastic dominance definition
$\int_0^x F(T)dT\le \int_0^x G(T)dT \iff \int u(x) dF(x) \ge \int u(x) dF(x)$
The solution is as follows:
But I cannot reach from this integral
$$\int_0^1\int_0^xF(z)dzdx\le \int_0^1\int_0^xG(z)dzdx$$
to this integral
$U(F)= \int u(x)dF(x) \ge U(G)= \int u(x)dG(x)$
How can I get this last integral from the integral in yellow box? I asked this way.
A MPS ia such that $S(x) =\int^x F(z) dz < \int^x G(z) dz T(x) $. Recall that for any $u\in C^{\infty} $ you have the integration by parts formula
$$\int^b u(x) dF(x) = (boundaryterms)(u, F) - \int^b u'(x) F(x) dx = (boundaryterms) +\int^b u''(x) S(x) dx =(*) $$
If $u$ is concave $u''$ is always negative, so the inequality $S \le T$ yields
$$ (*) \ge (boundaryterms) (u, F) + \int^b u''(x) T(x) dx $$
To get the equality between boundary terms, you must use the boundary condition of MPS's, the endpoint of the interval you are working on. Using this and the same passages, you get
$$ (*) = \int^b u(x) dG(x) $$
Which conclude the proof.