If all trimmed means are equal does this imply equal distributions? I am trying to prove the following:
Given that $\forall \alpha\in [0,1]$:
$$\int_{F_S^{-1}(\alpha)}^{\infty}xf_S(x)\,dx = \int_{F_0^{-1}(\alpha)}^{\infty}yf_0(y)\,dy$$
where $F_S^{-1}(\alpha)$ and $F_0^{-1}(\alpha)$ are the $\alpha$th quantiles of the distribution. I wand to show that $f_S(x)$ is identically distributed to $f_0(x)$.
It seems like it should be true, but I haven't been able to prove it definitively. Any advice or directions to explore would be appreciated.
What I've tried so far:
integration by parts led me to the following:
$yF_{S}(y)\big|_{F_{S}^{-1}(\alpha)}^{\infty}-\int_{F_{S}^{-1}(\alpha)}^{\infty}F_{S}(y)dy=yF_{0}(y)\big|_{F_{0}^{-1}(\alpha)}^{\infty}-\int_{F_{0}^{-1}(\alpha)}^{\infty}F_{0}(y)dy$
But I can't substitute infinity for y in the first term without both sides going to infinity. To get around this I considered the fact that for some $\epsilon>0$:
$$\int_{F_S^{-1}(\alpha)}^{F_S^{-1}(\alpha+\epsilon)}xf_S(x)\,dx = \int_{F_0^{-1}(\alpha)}^{F_0^{-1}(\alpha+\epsilon)}yf_0(y)\,dy$$
which then applying integration by parts leads to:
$(\alpha+\epsilon)F_{S}^{-1}(\alpha+\epsilon)-\alpha F_{S}^{-1}(\alpha)-\int_{F_{S}^{-1}(\alpha)}^{F_{S}^{-1}(\alpha+\epsilon)}F_{S}(y)dy =
(\alpha+\epsilon)F_{0}^{-1}(\alpha+\epsilon)-\alpha F_{0}^{-1}(\alpha)-\int_{F_{0}^{-1}(\alpha)}^{F_{0}^{-1}(\alpha+\epsilon)}F_{0}(y)dy$
and I feel if I could show that the integral components of this were equal to 0 then I could solve it. 
 A: Recall that
$$\frac{\text{d}}{\text{d}\theta} \int_\theta^\infty f(x)\,\text{d}x=-f(\theta)$$
and hence that
$$\frac{\text{d}}{\text{d}\theta} \int_{g(\theta)}^\infty f(x)\,\text{d}x=-(f\circ g)(\theta)\times g'(\theta)$$
Hence, assuming the expectations of $F_S$ and $F_0$ are well-defined and finite,
$$\frac{\text{d}}{\text{d}\alpha}\int_{F_S^{-1}(\alpha)}^{\infty}xf_S(x)\,\text{d}x = - F_S^{-1}(\alpha)\times(f_S\circ F_S^{-1})(\alpha)\times\frac{\text{d}}{\text{d}\alpha}F_S^{-1}(\alpha)$$
Recall  further that
$$\frac{\text{d}}{\text{d}\alpha}F_S^{-1}(\alpha)=\frac{1}{[\frac{\text{d}}{\text{d}x}F_S](F_S^{-1}(\alpha))}=\frac{1}{f_S(F_S^{-1}(\alpha))}=\frac{1}{(f_S\circ F_S^{-1})(\alpha)}$$
and you should directly deduce the result.
This resolution is in fact equivalent to a change of variable in the integral$$\int_{F_S^{-1}(\alpha)}^{\infty}xf_S(x)\,\text{d}x$$
since expressing $x$ as $x=F_S^{-1}(\beta)$ in this integral leads to
$$\int_{\alpha}^{1}F_S^{-1}(\beta)\,\text{d}\beta$$thanks to the same cancellation of $(f_S\circ F_S^{-1})(\alpha)$ as above.
Hence
$$\int_{\alpha}^{1}F_S^{-1}(\beta)\,\text{d}\beta=\int_{\alpha}^{1}F_0^{-1}(\beta)\,\text{d}\beta$$
for all $0<\alpha<1$ and taking the derivative in $\alpha$:
$$-F_S^{-1}(\alpha)=F_S^{-1}(\alpha)$$
for all $0<\alpha<1$, as expected.
