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Is it true that in the Riesz representation theorem

$\mu(F)=\sup\{\Lambda(f): f\in C_c, 0\leq f \leq 1, \operatorname{supp} f \subset F \}$

for every compact (or closed) subset $F$?

(It is known that it holds if $F$ is open, and that $\mu(F)=\inf\{\Lambda(f): f\in C_c, 0\leq f \leq 1, f(x)=1 \ for \ x \in F \}$ if $F$ is compact.) I think it may be true, because in "Abstract harmonic analysis", vol.I., by E. Hewitt and K. Ross (if I well understood) the measure in the representation theorem is defined for closed subsets just in such a way (chap.11, (def. 11.20, th. 11.17 (i), def.11.11).

Thanks

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Try a fat Cantor set $F$. Note that in order for there to be a non-zero continuous $f \geq 0$ supported inside $F$ the set $F$ must have non-empty interior. In other words: no it's not true. –  t.b. Aug 8 '11 at 8:22
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up vote 2 down vote accepted

Not in general. Consider the example $\Lambda(f)=f(x)$ for some non-isolated point $x$, and $F=\{x\}$. More generally, if there are compact sets with positive measure and dense complement, that equality will not hold. Even for Lebesgue measure these exist, fat Cantor sets being one type of example.

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Maybe it's worth pointing out the general reflex one should have when asking such questions: Is the desired result true for Lebesgue measure on $[0,1]$? Is it true for a point measure? What about a sum of a point measure and Lebesgue measure, then? Most of the questions the OP had so far could be decided by following this algorithm. –  t.b. Aug 8 '11 at 8:40
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