# $\mu$ and $\nu$ be Borel measures on $\mathcal{B}(X), \mu << \nu$ then $\mu$ is Radon Measure

Let $\mu$ and $\nu$ be Borel measures on $\mathcal{B}(X)$, where $X$ is a compact topological space, and suppose that $\mu$ is absolutely continuous with respect to $\nu$. If $\nu$ is Radon show that $\mu$ also is Radon.

I need some help connecting all the pieces together:

$\mu$ and $\nu$ Borel Measures : provided every compact subset of $X$ has finite measure.

$\mu$ is absolutely continuous with respect to $\nu$ : if $E \in \mathcal{B}(X)$ s.t. $\nu(E)=0$ then $\mu(E)=0$

$\nu$ is Radon measures :

1. $\nu$ is Borel measure
2. (Outer Regularity) for each Borel subset $E$ of $X$ $\nu(E) =\inf \{ \nu(\mathcal{U} \ | \ \mathcal{U} \text{ a neighborhood of } E\}$
3. (Inner Regularity) for each open subset $\mathcal{O}$ of $X$ $\nu(\mathcal{O} )= \sup \{ \nu(K) \ | \ K \text{ a compact subset of } \mathcal{O}\}$

We want to prove that: $\mu$ is Radon measures :

1. $\mu$ is Borel measure
2. (Outer Regularity) for each Borel subset $E$ of $X$ $\mu(E) =\inf \{ \mu(\mathcal{U} \ | \ \mathcal{U} \text{ a neighborhood of } E\}$
3. (Inner Regularity) for each open subset $\mathcal{O}$ of $X$ $\mu(\mathcal{O} )= \sup \{ \mu(K) \ | \ K \text{ a compact subset of } \mathcal{O}\}$

Any help is appreciated. Thanks

Post edited, as original post was a bit of a mess.

Which definition of Radon measure are you using? At least according to wikipedia, a Radon measure should also be locally finite.

Partial answer: To prove $\mu$ is Radon measure, we have to prove $\mu$ is locally finite, inner regular and outer regular. In this partial answer, we will assume $\mu$ is locally finite, and using this, we will prove it is inner and outer regular.

Since we have assumed $\mu$ is locally finite, and $X$ is compact, we deduce $\mu$ is finite. Since $\nu$ is a Radon measure, it is also locally finite, hence it is also finite.

Outer regularity: To prove this, we can consider this characterization of absolute continuity instead:

If $\mu, \nu$ are finite, then $$\mu\ll \nu\iff \forall \epsilon>0\ \exists \delta>0\ s.t. \nu(A)<\delta \implies \mu(A)<\epsilon$$

Now we will show the outer regularity:

Assume by contradiction there is a Borel subset $E\subset X$ s.t. $$\mu(E)<\inf\{\mu(\mathcal{U})|\mathcal{U}\textrm{ neighbourhood of }E \}:=I$$ Let $\epsilon =\frac{I-\mu(E)}{2}$. Then, for all $\delta>0$, there is a neighbourhood $\mathcal{U}$ of $E$ s.t. $\nu(\mathcal{U}\setminus E)<\delta$, yet $\mu(\mathcal{U}\setminus E)>\epsilon$. This contradicts $\mu\ll \nu$.

Inner regularity:

Assume by contradiction there is a Borel subset $E\subset X$ s.t. $$\mu(E)>\sup\{\mu(K)|K\textrm{ compact subset of }E \}:=S$$ Let $\epsilon =\frac{\mu(E)-S}{2}$. Then, for all $\delta>0$, there is a compact subset $K$ of $E$ s.t. $\nu(E\setminus K)<\delta$, yet $\mu(E\setminus K)>\epsilon$. This contradicts $\mu\ll \nu$.

• Good question, it does not mention anything about Borel measure been finite – Lucas Apr 3 '16 at 3:16
• you mean inner regularity should be similar, what you just proves is outer regularity, do we still prove inner regularity by contradiction? – Lucas Apr 3 '16 at 3:28
• Of course, you are right. My bad. Btw I've edited the answer, though I'm afraid it is not very useful. If it is found not very useful I will delete it. – Nate River Apr 3 '16 at 14:20
• The general assumption we assume $(X, \mathcal{T})$ is a Hausdorff space, I don't see anything wrong with your answer, i still have difficulty finding a contradiction for inner regularity, if you can give me a tip for inner regularity, it will be greatly apreciated – Lucas Apr 3 '16 at 15:12