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A certain functional $T$ is defined as: $$T(F)=\int_{(0,1)}F^{-1}(s)M(ds)$$ where $M$ is a probability measure with support $[\alpha,1-\alpha]$,for $\alpha>0$. The result that above functional is continuous at $F$, if the measure $M$ does not assign mass to the point where $F^{-1}$ is discontinuous is special case of Theorem 3.7. Robust Statistics by Huber and Ronchetti(2nd ed).

If $\frak{M}$ be a set of probability measures, is there a way to guarantee that $$T_{sup}(F)=\sup_{\{M\in\frak{M}\}}\{\int_{(0,1)}F^{-1}(s)M(ds)\}$$ is continuous if none of $M$ assign mass to point where $F^{-1}$ is discontinuous? Is it false?

We can at least say that it is lower semi continuous. Can we say more?

Details Added:

Domain of $T$ is CDFs. $F^{-1}(x):=\inf\{y\in\mathbb{R}:F(y)\ge x\}$

Edit

The integral should be $$T_{sup}(F)=\sup_{\{M\in\frak{M}\}}\{\int_{[\alpha,1-\alpha]}F^{-1}(s)M(ds)\}$$ for $\alpha>0$

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  • $\begingroup$ The question is not clear. What is the domain of $T$? Do you mean by $F^{-1}(s)$ the multiplicative inverse $1/F(s)$? $\endgroup$
    – Jochen Wengenroth
    Commented Nov 25, 2014 at 15:05
  • $\begingroup$ @JochenWengenroth, Sorry, I avoided details as I thought the details might not effect the result. I have added details if that can help. $\endgroup$
    – Dinesh
    Commented Nov 25, 2014 at 16:05
  • $\begingroup$ When you say that the domain are 'random variables', do you mean 'cumulative distribution functions'? What topology do you want on these? (Presumably weak convergence for the corresponding probability measures?) $\endgroup$
    – Martin Hairer
    Commented Nov 25, 2014 at 21:48
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    $\begingroup$ You'll certainly need some additional assumptions since $F^{-1}$ typically diverges near $0$ and $1$, so that $T$ can easily be infinite. This is extremely unstable under weak convergence of the CDF $F$, so that $T$ cannot be continuous if $M$ has support at $0$ or at $1$. $\endgroup$
    – Martin Hairer
    Commented Nov 26, 2014 at 10:39
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    $\begingroup$ @MartinHairer, thank you very much for your remark. That was a mistake, sorry. I am in fact looking at the integral to $T_{sup}(F)=\sup_{\{M\in\frak{M}\}}\{\int_{[\alpha,1-\alpha]}F^{-1}(s)M(ds)\}$ for $\alpha>0$ $\endgroup$
    – Dinesh
    Commented Nov 26, 2014 at 11:19

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If all you assume is that none of the $M$ assigns mass to a point where $F^{-1}$ is discontinuous then the statement is certainly false. Assume you have a discontinuity at ${1/2}$, it suffices to consider the case where $\mathfrak{M}$ consists of a sequence of Dirac measures located at points that accumulate at $1/2$ from below...

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