# If $F$ is closed, does there exist a finite measure on $[0,1]$ with support $F$?

If $X$ is a metric space, $\mathcal B$ is the Borel $\sigma$-algebra, and $\mu$ is a measure on $(X, \mathcal B)$, then the $support$ of $\mu$ is the smallest closed set $F$ such that $\mu(F^c) = 0$. Show that if $F$ is a closed subset of $[0,1]$, then there exists a finite measure on $[0,1]$ whose support is $F$.

My Solution:

Let $F$ be a closed subset of $[0,1]$ and $\mu_F$ be defined by $\mu_F(B) = m(B \cap F)$ for any $B \in \mathcal B_{[0,1]}$. Since $F$ is a closed subset of $[0,1]$, it is in $\mathcal B_{[0,1]}$ and $\mu_F$ is a measure. The measure is finite since $\mu_F([0,1]) = m([0,1] \cap F) \leq m([0,1]) = 1$ and since $\mu_F(F^c) = m(F^c\cap F) = m(\emptyset) = 0$, the support of $\mu_F$ is $F$.

My questions:

Why do we need the condition that $X$ is a metric space to define the support?

• For your first question: mathoverflow.net/questions/44408 – user940 Jun 7 '14 at 16:55
• – user940 Jun 7 '14 at 16:57
• As for your answer, you still need to prove that $F$ is the smallest such closed set. – user940 Jun 7 '14 at 17:24
• @ByronSchmuland: Which will fail. Suppose for instance that $F$ is a single point, or any other nonempty closed set with Lebesgue measure zero. Then your $\mu_F$ is the zero measure, whose support is $\emptyset$, not $F$. – Nate Eldredge Jun 7 '14 at 17:39
• math.stackexchange.com/a/208308/21674 – Michael Greinecker Jun 7 '14 at 19:41

Second Try:

Let $F$ be a closed subset of $[0,1]$. Since any subset of $[0,1]$ is separable, there exists a countable set $A \subset F$ such that $F$ is the closure of $A$.

If $A = \{a_1, a_2,\dots, a_k\}$ is finite, let $\mu(B) = \frac{1}{k}\cdot n(B \cap A)$ where $n$ is the counting measure. Then $\mu([0,1]) = \frac{1}{k}\cdot n([0,1] \cap A) = \frac{1}{k}\cdot n(A) = 1$ so that $\mu$ is finite, and $\mu(F^c) = \frac{1}{k}\cdot n(F^c \cap A) = \frac{1}{k}\cdot n(\emptyset) = 0$. Since $F$ is the closure of $A$, it is the smallest closed set containing $A$ and thus is the smallest closed set such that $F^c$ has measure zero. So, $F$ is the support of $\mu$.

If $A = \{a_1, a_2, \dots\}$ is infinite, let $\nu(B) = \sum_{n=1}^\infty \frac{1}{2^n}\cdot \delta_{a_n}(B)$ where $\delta_n$ is the dirac-$\delta$ measure. Since the countable sum of measures is a measure, $\nu$ is a measure. Furthermore, it is finite since $\nu([0,1]) = \sum_{n=1}^\infty \frac{1}{2^n}\cdot \delta_{a_n}([0,1]) =\sum_{n=1}^\infty \frac{1}{2^n} = 1$ (because $\delta_{a_n}([0,1]) = 1$ for all $n \in \mathbb N$). Also, since $A \subset F$, $\delta_{a_n}(F^c) = 0$ for all $n \in \mathbb N$, and $\nu(F^c) = \sum_{n=1}^\infty \frac{1}{2^n}\cdot \delta_{a_n}(F^c) = 0$. Again, since $F$ is the closure of $A$, it is the smallest closed set such that $F^c$ has measure zero. So, $F$ is the support of $\nu$.

In any case, there exists a finite measure on $[0,1]$ whose support is $F$.

• Why is $F$ is smallest closed set with $\nu(F^c)=0$? – user940 Jun 8 '14 at 0:01
• I should say, if there were a proper closed subset of $F$, it could not possibly contain $A$ since this would contradict $F$ being the closure. Thus, its complement intersects $A$, giving it positive measure. – dannum Jun 8 '14 at 0:26

To expand on danielson's proof, it still remains to check that $F$ is the smallest closed set such that $\nu(F^c) = 0$. Let $G \subset F$ be closed and a strict subset of $F$, so there is a point $x \in F$ but $x \not\in G$. Since $x \in G^c$, which is an open set, there is a $\delta > 0$ such that $B(x, \delta) \subset G^c$. Now, put $\delta_1 := \delta$ and let $\delta_n \searrow 0$. There are two cases: either (a) there exists a $0 < \delta_n \leq \delta$ such that $B(x, \delta_n) \subset F$, or (b) $x$ is an isolated point of $F$. In case (a), there is some $a_i \in B(x, \delta_n)$ by the denseness. In case (b), $x$ must be a member of the dense subset of $F$. In either case we have shown that $\nu(G^c) > 0$.