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?
Is my answer correct?
 A: 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 $F$.
A: 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 $\nu(G^c) > 0$. 
A: Here is my attempt to your problem.
Define the following measure $\mu_{F}(A)$\begin{cases}
\begin{array}{c}
\mu(A)\\
0
\end{array} & \begin{array}{c}
A\cap F\not=\emptyset\\
A\cap F=\emptyset
\end{array}\end{cases}
where $A\in\mathcal{A} (\sigma-algebra)$ and $\mu$ is the Lebesgue measure. 
This is a measure since $\mu_{F}(\emptyset)=0$ and $\mu_{F}(\bigcup A_{i})=\mu(\bigcup A_{n_{i}})=\sum\mu(A_{n_{i}})=\sum\mu_{F}(A_{n_{i}})=\sum\mu_{F}(A_{i})$, the last equality is because I am adding measure zero sets. 
I do not have a lot of time right now and I will get back to this later but for this measure $F$ is the smallest closed set and of course $\mu_{F}([0,1])=\mu([0,1])=1$. 
I'd appreciate if you give me feedback on this since I just started learning measure theory.
A: Your solution is not  true. Note that if $F=\{x_1,x_2\}$, with $x_1,x_2\in[0,1]$, $\mu_F({\{x_1\}^c})=\mu(\{x_2\})=0$. As $\{x_1\}$ is closed and subset of $F$, $F$ is not the support if $\mu_F$.
The solution of dannum is better.
