 # Consider the measure &#x03BC;<!-- μ --> on R such that &#x03BC;<!-- μ --> ( [ glycleWogry 2022-06-25 Answered
Consider the measure $\mu$ on $R$ such that $\mu \left(\left[-r,r\right]\right)>0$ 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?
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I suppose you want $f$ to be a non-negative smooth function.
If $\left({r}_{n}\right)$ 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 ${\int }_{E}f\left(x\right)dx\le m\left(E\right)=0$ 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 $\mu \left(E\right)\ge {\int }_{E}f=0$ for all $E$ in side ${S}^{c}$ and ${\int }_{E}f\left(x\right)xd=0$ for $E$ inside $S$ also. So $f=0$ a.e. [Lebesgue].

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