# Integral defines a Radon measure

Suppose that $\mu$ is a Radon measure on $X$. If $\phi\in L^{1}(\mu)$ and $\phi\geq 0$, show that $\nu(E)=\int_{E}\phi\;d\mu$ is a Radon measure.

$\nu$ is a finite measure so we just need to check the regularity conditions. I was showing the outer regularity on all Borel sets and I did it for sets with finite $\mu$-measure. I'm not sure about the argument for sets with infinite $\mu$-measure. Likewise for inner regularity on open sets.

• I probably should add that $X$ is a locally compact Hausdorff space. – cyc Apr 13 '13 at 16:36
• Presumably not $\sigma$-finite, otherwise you would have finished already... – copper.hat Apr 13 '13 at 17:00
• @copper.hat: Yes indeed. – cyc Apr 13 '13 at 17:36
• It probably helps to note that $\nu$ vanishes outside the $\sigma$-finite set $U = \bigcup_{n} \{ \varphi \gt 1/n\}$, so you can reduce to the case that $\mu$ is $\sigma$-finite. – Martin Apr 13 '13 at 18:56
• @Martin: I'm not sure how the reduction goes. – cyc Apr 13 '13 at 21:13

This is just an elaboration of Martin's astute comment above:

Let $N = \phi^{-1} \{0 \}$. Then $\nu A = \nu (A \setminus N)$. Furthermore, $N^c = \cup_k \Delta_k$, where $\Delta_k = \phi^{-1} (\frac{1}{k}, \infty)$.

We can bound $\mu \Delta_k$ as follows: $\|\phi\|_1 \ge \nu \Delta_k = \int_{\Delta_k} \phi d \mu \ge \frac{1}{k} \mu \Delta_k$, and so $\mu \Delta_k \le k \|\phi\|_1$.

We have $A \setminus N = \cup_k (A \cap \Delta_k)$, and since $\Delta_k$ is increasing, we have $\nu (A \cap \Delta_n) \to \nu(\cup_k (A \cap \Delta_k))$.

Let $\epsilon>0$, then choose $n$ such that $\nu(\cup_k (A \cap \Delta_k)) - \nu (A \cap \Delta_n)< \frac{\epsilon}{2}$.

Now we need to approximate $A \cap \Delta_n$ by a compact set. Since $\mu$ is a Radon measure, there exists a sequence of compact sets $C_k \subset A \cap \Delta_n$ such that $\mu C_k \to \mu (A \cap \Delta_n)$. Without loss of generality (finite unions of compact sets are still compact) we may assume that the $C_k$ are increasing. We have $\mu((A \cap \Delta_n) \setminus \cup_k C_k) = 0$, hence $1_{C_k}(x) \to 1_{A \cap \Delta_n}(x)$ a.e. [$\mu$]. Since $1_{C_k}(x) \le 1$, the dominated convergence theorem gives $\int_{C_k} \phi d \mu \to \int_{A \cap \Delta_n} \phi d \mu$. Hence for some $k$, we have $\int_{A \cap \Delta_n} \phi d \mu - \int_{C_k} \phi d \mu < \frac{\epsilon}{2}$, and so \begin{eqnarray} \nu A &=& \nu (A \setminus N) \\ &=& \nu ( \cup_k (A \cap \Delta_k)) \\ &<& \nu (A \cap \Delta_n) + \frac{\epsilon}{2} \\ &=& \int_{A \cap \Delta_n} \phi d \mu + \frac{\epsilon}{2} \\ &<& \int_{C_k} \phi d \mu + \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ &=& \nu C_k + \epsilon \end{eqnarray} Hence $\nu A = \sup \{ \nu C | C \subset A, C \text{ compact} \}$. $\nu$ is bounded by $\nu X = \|\phi\|_1$, hence is locally finite, and so $\nu$ is a Radon measure.

If you want to get outer regularity, let $A$ be a Borel set, and let $C_k \subset A^c$ be a sequence of compact sets such that $\nu C_k \to \nu A^c$. Now let $U_k = C_k^c$. Then $U_k$ is open and $A \subset U_k$. Since $\nu$ is finite, we have $\nu X = \nu A + \nu A^c = \nu U_k + \nu C_k$. It follows that $\nu U_k \to \nu A$. It follows that $\nu A = \inf \{ \nu U | A \subset U, U \text{ open} \}$, hence $\nu$ is outer regular.

• That settles the problem with inner regularity right? I'm still unsure about outer regularity. – cyc Apr 13 '13 at 21:37
• To be more precise, I want to arrive at $\nu(E)=\inf\{\nu(U):E\subset U,U\;\text{open}\}$ for all Borel sets $E$, and $\nu(V)=\sup\{\nu(K):K\subset V,K\;\text{compact}\}$ for all open sets $V$. I can see that any set $E$ can be partitioned into two parts by $N=\phi^{-1}\{0\}$. One part is $\sigma$-finite which respect to $\mu$ and the other is $\nu$-null. For outer regularity, I can probably find an open set approximating the $\sigma$-finite part, but I need an open set that contains the whole of $E$ and that's bothering me. – cyc Apr 14 '13 at 1:02
• @learner: I added something about outer regular to the answer. – copper.hat Apr 14 '13 at 2:14
• What downvote? Did something happen that I'm not aware of? – cyc Jul 2 '14 at 6:19
• @learner: I got a downvote about 6 hours ago. I don't care about the downvote other than understanding what was wrong. It is not particularly helpful if the downvoter doesn't indicate why. – copper.hat Jul 2 '14 at 6:32

Suppose $\{E_n : n \geq 1\}$ is a maximal family of pairwise disjoint positive measure compact subsets of $X$. Then $Y = X \backslash \bigcup \{E_n : n \geq 1\}$ must be $\mu$-null otherwise we can add a positive measure compact subset of $Y$ to our family. Doesn't this show that $\mu$ must be sigma finite?

Edit: This is flawed in that such a maximal family may be uncountable.

• I'm not sure I fully understand. How do you get this maximal family? – cyc Apr 13 '13 at 18:44
• Consider the collection of all families of pairwise disjoint compact subsets of X of positive measure. This collection is clearly closed under increasing unions hence we can apply Zorn's lemma. Oh I see the problem, it maybe possible that this maximal family is uncountable - Bizarre! – hot_queen Apr 13 '13 at 18:51