# 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 Jul 24, 2015 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 Jul 25, 2015 at 1:18
• And one more thing: truly awesome solution well clear and written beautifully thanks @triple_sec Jul 25, 2015 at 1:20
• @zbigniew2015 You're very welcome! I'm glad to hear you found it helpful. Jul 25, 2015 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. Jul 25, 2015 at 17:14
• @PhoemueX: Thank you got it very helpful for a smoother proof Jul 25, 2015 at 20:36