glycleWogry

2022-06-25

Consider the measure $\mu $ on $R$ such that $\mu ([-r,r])>0$ for all $r>0$.

Can we construct a (smooth) function $f$ satisfying $\mu -fdx\ge 0$ in a measure sense? If $\mu $ has a continuous density $g$, then it seems easy. But what conditions are needed for the existence of $f$ for a measure $\mu $? or is it possible always?

Can we construct a (smooth) function $f$ satisfying $\mu -fdx\ge 0$ in a measure sense? If $\mu $ has a continuous density $g$, then it seems easy. But what conditions are needed for the existence of $f$ for a measure $\mu $? or is it possible always?

popman14ee

Beginner2022-06-26Added 19 answers

I suppose you want $f$ to be a non-negative smooth function.

If $({r}_{n})$ is an evaluation of rational numbers and $\mu =\sum \frac{1}{{2}^{n}}{\delta}_{{r}_{n}}$ then the only function $f$ satsfying your conditioin is $f$=0 (up to a Lebesgue null set). This becasue ${\int}_{E}f(x)dx\le m(E)=0$ for any Borel set $E$ contained in $\mathbb{R}\setminus \mathbb{Q}$ and this implies ${\int}_{E}f=0$ for all Borel sets $E$.

Note: If a smooth function is 0 a.e. [Lebesgue] then it is 0 at every point.

The same conclusin holds for any $\mu $ singualr w.r.t. Lebesgue measure. If $\mu $ is concentared on $S$ and $m$ (Lebesgue measure) on it complement then $\mu (E)\ge {\int}_{E}f=0$ for all $E$ in side ${S}^{c}$ and ${\int}_{E}f(x)xd=0$ for $E$ inside $S$ also. So $f=0$ a.e. [Lebesgue].

If $({r}_{n})$ is an evaluation of rational numbers and $\mu =\sum \frac{1}{{2}^{n}}{\delta}_{{r}_{n}}$ then the only function $f$ satsfying your conditioin is $f$=0 (up to a Lebesgue null set). This becasue ${\int}_{E}f(x)dx\le m(E)=0$ for any Borel set $E$ contained in $\mathbb{R}\setminus \mathbb{Q}$ and this implies ${\int}_{E}f=0$ for all Borel sets $E$.

Note: If a smooth function is 0 a.e. [Lebesgue] then it is 0 at every point.

The same conclusin holds for any $\mu $ singualr w.r.t. Lebesgue measure. If $\mu $ is concentared on $S$ and $m$ (Lebesgue measure) on it complement then $\mu (E)\ge {\int}_{E}f=0$ for all $E$ in side ${S}^{c}$ and ${\int}_{E}f(x)xd=0$ for $E$ inside $S$ also. So $f=0$ a.e. [Lebesgue].

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