Support of a Lebesgue-Stieltjes measure

Question

Let $G:[0,1]\to[0,1]$ be a distribution function (that is, non-decreasing and right-continuous) with support $\text{supp}(G)$ define as:

$$\text{supp}(G)=\{r\in [0,1]: G(r+\varepsilon)-G(r-\varepsilon)>0\;\forall \varepsilon>0\}.$$

Associated with $G$, there exists a Lebesgue-Stieltjes probability measure $\mu_{G}$ s.t. for $I=(a,b]$ with $0<a<b\le 1$, $\mu_{G}(I)=G(b)-G(a)$. The support of $\mu_{G}$ is the smallest closed set with measure 1.

1. Show that $\text{supp}(\mu_{G})=\text{supp}(G)$.
2. If $S\in[0,1]\backslash \text{supp}(G)$, then $\mu_{G}(S)=0$, but the converse not necessarily true. Under what conditions does the converse hold?

Answer 1

For Item 1:

For brevity, let $S_1 = \operatorname{supp}(\mu_G)$, $S_2 = \operatorname{supp}(G)$, and $S_3 = \{r\in[0,1] : \forall U\text{ open in } [0, 1] \text{ with } r\in U \implies \mu_G(U) > 0 \}$.

1. I will show $S_1 = S_3$ first: Notice that $S_1$ is closed and $\mu_G(S_1) = 1$. Let $r\in S_1$ and $U$ open with $r\in U$. Then, $S_1\setminus U$ is closed and thus we have $\mu_G(S_1 \setminus U) < 1$. In particular, we have $$\mu_G(U) \ge \mu_G(S_1 \cap U) = 1 - \mu_G(S_1 \setminus U) > 0.$$ On other hand, let $r\in S_3$ and $C$ be closed with $\mu_G(C) = 1$. Assume $r\not\in C$. Then, there exists an open neighborhood $U$ of $r$ disjoint to $C$. But that implies $$ \mu_G(C \cup U) = \mu_G(C) + \mu_G(U) > 1,$$ which contradicts that $\mu_G([0, 1]) = 1$.

2. Proof for $S_2\subseteq S_3$: Let $r\in S_2$ and $\varepsilon > 0$. Then, we have $$ 0 < \mu_G((r - \varepsilon, r + \varepsilon)) \le \mu_G((r - \varepsilon, r + \varepsilon]) = G(r+\varepsilon) - G(r - \varepsilon) $$ and therefore $r\in S_3$.

3. Proof for $S_3\subseteq S_2$: Now, let $r\in S_3$, $U$ be open with $r\in U$, and $\varepsilon > 0$ with $(r-\varepsilon, r+\varepsilon) \subseteq U$. Then, we have $$ 0 < G(r+\varepsilon/2) - G(r - \varepsilon/2) = \mu_G((r - \varepsilon / 2, r + \varepsilon / 2]) \le \mu_G((r - \varepsilon, r + \varepsilon)) \le \mu_G(U) $$ and therefore $r\in S_2$.

For Item 2:

The question is a little bit vague. It is surely sufficient that $S$ has non-empty interior.