# Borel Measure

#### Fermat

##### Active member
Let K be a compact hausdorff space, and u a borel measure on K. You are given that if A is an open set in K with A and E disjoint, we have u(A)=0. (E is a certain closed set in K)
Show that for a borel set A, we have that u(AE)=u(A), where AE is the intersection.

we have that u(A)=u(AE)+u(A\E) so we only have to prove u(A/E)=0

u(A/E)=inf(u(M):M is open and A\E is a subset of M)

A\E is disjoint with E but unfortunately this does not imply M is disjoint with E .

#### Opalg

##### MHB Oldtimer
Staff member
Let K be a compact hausdorff space, and u a borel measure on K. You are given that if A is an open set in K with A and E disjoint, we have u(A)=0. (E is a certain closed set in K)
Show that for a borel set A, we have that u(AE)=u(A), where AE is the intersection.

we have that u(A)=u(AE)+u(A\E) so we only have to prove u(A/E)=0

u(A/E)=inf(u(M):M is open and A\E is a subset of M)

A\E is disjoint with E but unfortunately this does not imply M is disjoint with E .
This argument seems too simple to be believable, but I can't see what's wrong with it (I'm suspicious, because it does not use the compactness of $K$). You are given that $E$ is closed, so its complement $K\setminus E$ is an open set disjoint from $E$. Therefore $u(K\setminus E) = 0.$ If $A$ is a Borel set then $u(A\setminus E) \leqslant u(K\setminus E) = 0$.

#### Fermat

##### Active member
Is the last part simply because A\E is a subset of K\E?

#### Opalg

##### MHB Oldtimer
Staff member
Is the last part simply because A\E is a subset of K\E?
Yes. If $X$, $Y$ are Borel sets with $X\subseteq Y$ then $u(X)\leqslant u(Y)$.

#### girdav

##### Member
Indeed, in the use of outer regularity, there is a problem as the involved $M$ may not be disjoint with $E$.

We have to show that the measures $\mu$ and $\nu\colon A\mapsto \mu(A\cap E)$ are actually equal.

Since the ambient space is compact, we can assume that $\mu$ is finite. We only have to check that the relationship $\mu(A)=\mu(A\cap E)$ holds for $A$ an open set. Using closeness of $E$, $A\cap E^c$ is an open set disjoint with $E$.

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