I’m trying to solve this exercise, but I have some doubts: I think point (b) is correct, but I’m not sure about point (a).

Let X be a locally compact Hausdorff topological space and let µ: S → [0, +∞] be a positive Radon measure on X.

Let $T = bigcup{V : V text{ is open and }mu(V) = 0}$

and $text{supp}(mu)=T^C$.

Prove that:

a) T is open and $ mu(T)=0 $

b) $ x in text{supp}(mu) iff int_X phi dmu >0 $ ∀ $ phi in C_c(X) $ | ${x} prec text{supp}(phi)$.

a) T is open because it is a union of open sets.

How do I prove that it has measure zero?

Initially, I thought of using the inner regularity on open sets and approximating the measure of T with the measures of compact sets contained in T and then, by compactness, extracting a finite subcover of open sets.

However, I do not think I can do this because I cannot say that T contains compact sets.

Perhaps I could argue by contradiction, so assume $ mu(T)=c>0$ and consider $x in T$.

If I could show that $x in text{supp}(mu)$, then I would have a contradiction, but I do not know how to do this.

b) $ (Leftarrow)$

Assume the thesis does not hold, i.e., $ exists phi in C_c(X) $ with ${x} prec text{supp}(phi) $| $ int_X phi dmu =0 $.

By continuity, since $phi(x)=1$, there exists an open set V containing x on which $phi$ is non-zero, so $ 0=int_X phi dmu=0 Rightarrow mu(V)=0$.

Since V is open, it follows that $ V in T $ and therefore $x in T$.

Then, $x notin text{supp}(mu) =T^C$, which contradicts the assumption.

$ (Rightarrow)$

Assume the thesis does not hold, i.e., $x notin text{supp}(mu)=T^C $.

Then $x in T$ and by definition of T, $ exists $ an open set $ V $ with $ x in V $ and $mu(V)=0$.

Since X is LCH, ${x}$ is compact and $V$ is open with ${x} subset V$, I can use the Urysohn lemma for LCH spaces and find $ phi in C_c(X) $ such that $ {x} prec text{supp}(phi) prec V$.

Then $ int_X phi dmu = int_V phi dmu + int_{Xsetminus V} phi dmu = 0 $ because $ mu(V) = 0 $ and $ text{supp}(phi) subset V$.

Therefore, if I negate the hypothesis, the thesis is false.