Support of a measure Let $\mu$ be a regular Borel measure on compact subset $K \subseteq [0, \infty[$. Define
$$S:=\operatorname{support}(\mu) = \{x \in K: \mu(U) > 0 \mathrm{ \ for \ every \ open \ subset \ U \ containing \ x}\}.$$
Consider the identity map $z: K \to \mathbb{C}$. Do we have
$$\inf\{C > 0\mid \forall x \in S: x \le C\}= \inf\{C > 0\mid \mu\{x\in K: x > C\}=0\}?$$
Attempt:
If $x \le C$ for almost every $x \in K$ and there exists $x \in S$ with $x  > C$, then we can easily deduce a contradiction. Hence,
$$\{C > 0: \forall x \in S: x \le C\} \subseteq \{C > 0: \mu \{x \in K: x > C\}=0\}.$$
Taking the infinimum of both sides, we conclude that
$$\inf \{C > 0: \mu \{x \in K: x > C\}=0\} \le \inf\{C > 0: \forall x \in S: x \le C\}.$$
How can we show the converse inequality?
 A: Hint: If $\mu \{x \in K: x>C\}=0$ then $S \subseteq [0,C]$: Suppose $y >C$. The $(y-r,y-r)$ is an open set containing $y$ with $\mu (K \cap (y-r,y+r))=0$ if $0< r <y-C$. Hence, $y \notin S$.
A: A few basic observations:

*

*Any measure $\mu$ on a measurable set $F\subset\mathbb{R}$ may be consider as a measure on $\mathbb{R}$: $\mu(S)=\mu(A\cap F)$ for all measurable set $A$.


*The support $S$  of any regular measure $\mu$ on $\mathbb{R}$ is closed: if $x\notin S$, then there is a small open ball $U_x$ around $x$ such that $\mu(U_x)=0$. Then every point in $U_x$ is not in $S$, that is $U_x\subset S^c$ ans so $S^c$ is open.
In the OP, since by assumption $\mu$ is a measure concentrated on a compact set $K$ (i.e. $\mu(A)=0$ for any measurable set $A\subset K^c$), then $S\subset K$ and so, $S$ is compact too. The left-hand side of the "identity" the OP wishes to study is the same as the supremum of the support $S$ of $\mu$, that is
$$ s^*:=\inf\{C > 0\mid \forall x \in S: x \le C\}=\sup S$$
This means that for any $a<s^*<b$, $\mu((s^*,b))=0$ and
$$\mu\big((a,b)\big) =\mu\big((a,s^*])=\mu\big(\{x\in K: a<x\leq x^*\}\big)>0$$
As $S$ is closed (actually compact since $S\subset K$ and $S$ is closed), $s^*\in S$.  In turn, this implies that  $\{x\in K: x>s^*\}\subset S^c$ and so,
$$\mu\big(\{x\in K: x>s^*\}\big)=0$$
Finally, if $C>0$ is such that $\mu\big(\{x\in K: x>C\}\big)=0$, then  $C\geq s^*$. Otherwise, if $C<s*$, then
$$\mu\big(\{x\in K: x>C\}\big)\geq \mu(K\cap(C,s^*])>0$$
contradicting the fact that $\mu\big(\{x\in K: x>C\}\big)=0$.
Putting things together, you get that
$$s^*=\inf\{C > 0\mid \mu\{x\in K: x > C\}=0\}$$
