While studying measurable function, I want to know if the following statement is true. Let $\mu$ be a measure on $\mathbb R^n$ with a support with nonempty interior, $f$ polynomial function and $B$ a compact subset of $\mathbb R^n$ : $$A = \left\{\int_{\Omega} f \mathrm d\mu : \Omega \subset B \text{ $\mu$-measurable}\right\}$$ is convex. Can any one help me to prove or disprove this statement ?
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$\begingroup$ Is the measure assumed to be a Borel measure? $\endgroup$– S. DewarApr 13, 2018 at 1:47
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$\begingroup$ Can't one take a measure with discrete support? E.g. $\mu(H):=|H\cap\{0,1\}|$ on $\Bbb R$. $\endgroup$– BerciApr 13, 2018 at 1:50
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$\begingroup$ If you add this assumption could you prove it $\endgroup$– KrokiApr 13, 2018 at 1:51
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$\begingroup$ @Berci, I add a codition on $\mu$. $\endgroup$– KrokiApr 13, 2018 at 1:53
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