# 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?

• What does the identity map enter into your problem? Apr 27 at 15:10
• @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 $K$.
– user839372
Apr 27 at 15:44
• 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. Apr 27 at 15:53

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 $$0< r . Hence, $$y \notin S$$.

• Don't you mean $(y-r,y+r)$? And don't you need to intersect it with $K$?
– user839372
Apr 27 at 13:41

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 $$a, $$\mu((s^*,b))=0$$ and $$\mu\big((a,b)\big) =\mu\big((a,s^*])=\mu\big(\{x\in K: a0$$ As $$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, 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 $$s^*=\inf\{C > 0\mid \mu\{x\in K: x > C\}=0\}$$

• Why is $\mu((s^*,b)) = 0?$
– user839372
Apr 27 at 20:19
• because $(s^*,b)$ is outside the support $S$ Apr 27 at 20:33
• 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?
– user839372
Apr 27 at 20:34
• if there were at least one point $p\in s\cap (s^*,b)$, then to would be larger than $s^*$ contradicting the definition of $s^*$. Apr 27 at 20:34
• Basically my question is: if $U \cap S = \emptyset$, then $\mu(U) = 0$?
– user839372
Apr 27 at 20:36