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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.

cyc
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1 Answers1

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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.

copper.hat
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  • 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 01:02
  • @learner: I added something about outer regular to the answer. – copper.hat Apr 14 '13 at 02:14
  • What downvote? Did something happen that I'm not aware of? – cyc Jul 02 '14 at 06:19
  • @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 02 '14 at 06:32
  • @copper.hat I was the one who downvoted your question. This question is an exercise in Folland (1999) and this text defines Radon measures to be inner regular only on open sets, and not necessarily on all Borel sets. In your answer, $A\cap\Delta_n$ is not necessarily open, in which case your argument does not go through. In the meanwhile, I did some research, though, and realized that it is actually customary to define Radon measure to be inner regular on all Borel sets, in which case your argument is valid. (cont'd) – triple_sec Jul 02 '14 at 19:50
  • In the light of this ambiguity, I retracted my downvote. (I inserted spaces in your answer so that I can retract the vote, that's why it shows I edited it, although I did not modify anything). I should have explained this concern of mine to you earlier; I apologize for not having done so until now. – triple_sec Jul 02 '14 at 19:51
  • @triple_sec: No problem, thanks for taking the time to explain. – copper.hat Jul 02 '14 at 20:11
  • @copper.hat Upon further reflection, your argument does go through even if one sticks with a more restrictive definition of Radon measures (that is, one that says that such measures are inner regular on open sets). This is because $\mu(\Delta_k)\leq k|\phi|_1<\infty$ for all $k\in\mathbb N$, given that $\phi\in L^1(\mu)$. Moreover, Radon measures (restrictively defined) are actually inner regular on ($\sigma$-)finite Borel sets (Proposition 7.5 in Folland, 1999, p. 216). Once again, apologies for my confusion. I hope you will accept an upvote as a token of my repentance. :-) – triple_sec Jul 03 '14 at 00:00
  • @triple_sec: No need for either :-). – copper.hat Jul 03 '14 at 00:09