# Question 7.7 in measure theory on Radon measure from Folland's Real Analysis Second Edition

Hello all I was presented with this question from Folland's real analysis second edition on Radon measures which I am stuck on and so would really appreciate the help on. I m a novice in Radon measures especially in the concepts and abstractions so any help would be appreciated. It is problem #7 on page 220 It reads as follows: Just in case, the definition of Radon Measure used in the book is: A Radon measure on X is a Borel measure that is finite on all compact sets, outer regular on all Borel sets, and inner regular on all open sets. I also know for a fact that Radon measures are also inner regular on all their sigma finite sets. I have tried to attack the problem many times but to no avail as for some reason I cannot seem to incorporate the given assumption that X is sigma finite. Also the assumption is X is locally compact Hausdorff space. I would really appreciate the help Thanks

EDIT: I figured it is obviously finite on compact sets

Possible direction I have: I have studied in the Folland book that all $\sigma-finite$ Radon measures are regular therefore for all Borel sets A and E we have $\mu_A(E)=\mu(E \cap A)$ is the supremum on measure of all compact subsets in $E \cap A$. How do I move further? I need inner regularity on open sets E meaning the supremum on measures of compact subsets of E not $A \cap E$ as I have.

• I figured it is obviously finite on compact sets and outer regular on all Borel sets now all is missing is to show it is inner regular on all open sets – kroner Jul 24 '15 at 23:15

Let's check whether $\mu_A$ satisfies the conditions of being a Radon measure step by step.

1. $\mu_A$ is a measure. This is easy, I leave it to you.

2. $\mu_A$ is finite on compact sets. Let $K\subseteq X$ be compact. Then, $$\mu_A(K)=\mu(K\cap A)\leq \mu(K)<\infty,$$ given that $\mu$ is a Radon measure.

3. $\mu_A$ is outer regular on all Borel sets. Let $E\in\mathscr B_X$. We want to show that $$\mu_A(E)=\inf\{\mu_A(U)\,|\,E\subseteq U\text{ and U is open}\}.$$ The direction $\leq$ is easy to see, for if $E\subseteq U$ and $U$ is open, then $\mu_A(E)\leq\mu_A(U)$; now take infimum. The challenging part is the direction $\geq$. Fix $\varepsilon>0$. Since $\mu$ is $\sigma$-finite, there exists a sequence of $(F_n)_{n\in\mathbb N}$ of Borel-measurable sets such that $\mu(F_n)<\infty$ for all $n\in\mathbb N$ and $X=\bigcup_{n=1}^{\infty} F_n$. Without loss of generality, the $(F_n)_{n\in\mathbb N}$ can be taken to be disjoint. (Exercise: why?) Now, for any fixed $n\in\mathbb N$, one can find an open set $U_n\subseteq X$ such that

• $E\cap F_n\subseteq U_n$; and

• $\mu(U_n)< \mu(E\cap F_n)+\varepsilon/2^n<\infty$,

given that $\mu$ is outer regular and $\mu(F_n)<\infty$. Note that

\begin{align*} \mu(U_n\cap A)=&\,\mu(U_n\cap [E\cap F_n]^c\cap A)+\mu(U_n\cap E\cap F_n\cap A)\\ =&\,\mu(U_n\cap [E\cap F_n]^c\cap A)+\mu(E\cap F_n\cap A)\\ \leq&\,\mu(U_n\cap [E\cap F_n]^c)+\mu(E\cap F_n\cap A)\\ =&\,\mu(U_n)-\mu(E\cap F_n)+\mu(E\cap F_n\cap A)<\mu(E\cap F_n\cap A)+\varepsilon/2^n. \end{align*} Letting $U\equiv\bigcup_{n\in\mathbb N}U_n$, we have that $U$ is open, $$E=\bigcup_{n\in\mathbb N}E\cap F_n\subseteq\bigcup_{n\in\mathbb N} U_n=U,$$ and \begin{align*} \mu_A(U)=&\,\mu(U\cap A)\leq\sum_{n=1}^{\infty}\mu(U_n\cap A)\leq\sum_{n=1}^{\infty}\mu(E\cap F_n\cap A)+\sum_{n=1}^{\infty}\frac{\varepsilon}{2^n}\\ =&\,\mu\left(\bigcup_{n\in\mathbb N}E\cap F_n\cap A\right)+\varepsilon=\mu(E\cap A)+\varepsilon=\mu_A(E)+\varepsilon. \end{align*} Since $\varepsilon$ can be made arbitrarily small, it follows that $$\mu_A(E)\geq\inf\{\mu_A(U)\,|\,E\subseteq U\text{ and U is open}\}.$$

4. $\mu_A$ is inner regular on open sets. The claim is that for any open subset $U\subseteq X$, $$\mu_A(U)=\sup\{\mu_A(K)\,|\,K\subseteq U\text{ and K is compact}\}.$$ Again, the direction $\geq$ is clear. For the direction $\leq$, let's consider two cases.

Case I: $\mu(U\cap A)<\infty$. Fix $\varepsilon>0$. By the outer regularity of $\mu$, there exists some open set $V\subseteq X$ such that

• $U\cap A\subseteq V$; and
• $\mu(V)<\mu(U\cap A)+\varepsilon<\infty$.

Now, $U\cap V$ is an open set of finite measure. In turn, by the inner regularity of $\mu$, there exists some compact set $K\subseteq X$ such that

• $K\subseteq U\cap V$; and
• $\mu(K)>\mu(U\cap V)-\varepsilon$.

Therefore, \begin{align*} \mu_A(K)=&\,\mu(K\cap A)=\mu(U\cap V\cap A)-\mu(U\cap V\cap K^c\cap A)\\ =&\,\mu(U\cap A)-\mu(U\cap V\cap K^c)\\ =&\,\mu(U\cap A)-[\mu(U\cap V)-\mu(K)]\\ >&\,\mu(U\cap A)-\varepsilon=\mu_A(U)-\varepsilon, \end{align*} and also $K\subseteq U\cap V\subseteq U$. Therefore, $$\mu_A(U)\leq\sup\{\mu_A(K)\,|\,K\subseteq U\text{ and K is compact}\}.$$

Case II: $\mu(U\cap A)=\infty$. Fix any number $C>0$. Let $(F_n)_{n\in\mathbb N}$ be as before. Since $$\infty=\mu(U\cap A)=\sum_{n=1}^{\infty}\mu(U\cap A\cap F_n),$$ there must exist some $N\in\mathbb N$ such that $$C+1<\sum_{n=1}^{N}\mu(U\cap A\cap F_n)=\mu\left[U\cap A\cap\left(\bigcup_{n=1}^N F_n\right)\right]<\infty.$$ Let $G\equiv\bigcup_{n=1}^N F_n$. Again, one can find an open set $V\subseteq X$ such that

• $U\cap A\cap G\subseteq V$; and
• $\mu(V)<\mu(U\cap A\cap G)+1<\infty$.

This way, $U\cap V$ is an open set of finite measure. In turn, we can find a compact set $K\subseteq X$ such that

• $K\subseteq U\cap V\subseteq U$; and
• $\mu(K)>\mu(U\cap V)-1$.

We now find that \begin{align*} \mu_A(K)=&\,\mu(K\cap A)=\mu(U\cap V\cap A)-\mu(U\cap V\cap K^c\cap A)\\ \geq&\,\mu(U\cap A\cap G)-\mu(U\cap V\cap K^c)\\ =&\,\mu(U\cap A\cap G)-[\mu(U\cap V)-\mu(K)]\\ >&\,(C+1)-1=C. \end{align*} In words, if $\mu_A(U)=\infty$, then $U$ contains compact sets of arbitrarily large $\mu_A$-measure. The proof is complete.

• Thank you very much you truly helped me lots and lots. Really really appreciated – kroner Jul 25 '15 at 1:18
• And one more thing: truly awesome solution well clear and written beautifully thanks @triple_sec – kroner Jul 25 '15 at 1:20
• @zbigniew2015 You're very welcome! I'm glad to hear you found it helpful. – triple_sec Jul 25 '15 at 1:29
• @zbigniew2015: The proof of the inner regularity can actually be simplified considerably. As the OP writes, he knows that $\mu$ is inner regular on any $\sigma$-finite set (hence on any measurable set in the current situation). Now, let $E$ be arbitrary and $\alpha < \mu_A (E) = \mu(A \cap E)$. Then there is a compact set $K \subset A \cap E \subset E$ with $\mu(K) > \alpha$. But since $K \subset A$, this means $\mu_A (K) = \mu(A \cap K) = \mu(K) > \alpha$ and we are done. – PhoemueX Jul 25 '15 at 17:14
• @PhoemueX: Thank you got it very helpful for a smoother proof – kroner Jul 25 '15 at 20:36