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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?

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  • $\begingroup$ What does the identity map enter into your problem? $\endgroup$ Apr 27 at 15:10
  • $\begingroup$ @OliverDiaz It actually doesn't. But you can view the left side as the supnorm of the identity on $S$ and the right side as the $p=\infty$ norm of the identity on $K$. $\endgroup$
    – user839372
    Apr 27 at 15:44
  • $\begingroup$ I figured that much. Anyway, the left hand side of your target inequality is nthing but the supremum of the support of $S$. Also, to make things much simler, your measure $\mu$ can be thought of as a measure on $\mathbb{R}$ concentrated in $K$, for example $\mu(A)=\mu(A\cap K)$ for any measureble set $A$. That way, it is not important to consider sets of the form $\{x\in K:\ldots\}$. I hope my answer helps you understand this. $\endgroup$ Apr 27 at 15:53
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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, $y \notin S$.

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  • $\begingroup$ Don't you mean $(y-r,y+r)$? And don't you need to intersect it with $K$? $\endgroup$
    – user839372
    Apr 27 at 13:41
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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 $$s^*=\inf\{C > 0\mid \mu\{x\in K: x > C\}=0\}$$

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  • $\begingroup$ Why is $\mu((s^*,b)) = 0?$ $\endgroup$
    – user839372
    Apr 27 at 20:19
  • $\begingroup$ because $(s^*,b)$ is outside the support $S$ $\endgroup$ Apr 27 at 20:33
  • $\begingroup$ Yes, but it is still possible that a single open set outside the support has positive measure? The support are the points for which EVERY open neighborhood has positive measure. Am I missing something? $\endgroup$
    – user839372
    Apr 27 at 20:34
  • $\begingroup$ if there were at least one point $p\in s\cap (s^*,b)$, then to would be larger than $s^*$ contradicting the definition of $s^*$. $\endgroup$ Apr 27 at 20:34
  • $\begingroup$ Basically my question is: if $U \cap S = \emptyset$, then $\mu(U) = 0$? $\endgroup$
    – user839372
    Apr 27 at 20:36

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