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Let $X$ be locally compact Hausdorff space and $\Lambda$ be a positive linear functional on $C_c(X)$. It is known [W.Rudin, Real and complex analysis, th.2.14] that the measure $\mu$ in the Riesz representation theorem is given on an open set $V$ by formulae: $$\mu(V)=\sup \{\Lambda(f): f\in C_c, 0\leq f \leq 1, \operatorname{supp}{f} \subset V \}.$$ Is it true that $$\mu(V)=\sup\{\Lambda(f): f\in C_c, 0\leq f \leq 1, \operatorname{supp}{f} \subset \operatorname{cl}{ V} \}$$ for open $V$ ? Thanks.

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up vote 4 down vote accepted

No, this is not true. Far from it. Enumerate the rational numbers in $[0,1]$ by $q_1, q_2, q_3, \ldots$. For a given $0 \lt \varepsilon \lt 1$ put $U_{n} = [0,1] \cap (q_n - \varepsilon/2^{n+1},q_n + \varepsilon/2^{n+1})$ and observe that $V = \bigcup_{n=1}^{\infty} U_{n}$ is open and has Lebesgue measure at most $\varepsilon \gt 0$ by $\sigma$-subadditivity. In particular the first supremum is $\leq \varepsilon$. On the other hand, the closure of $V$ is $\overline{V} = [0,1]$ since $V$ is dense. Hence the second supremum is equal to $1$.

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Thank you very much for your help. – Richard Aug 6 '11 at 17:14

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