Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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 $1$.

share|improve this answer
Thank you very much for your help. –  Richard Aug 6 '11 at 17:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.