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Let $U\subseteq \mathbb{R}^d$ be an open set, and let $g:U\rightarrow [0,\infty)$ be a continuous map. It seems that the Borel measure defined by:

$\nu(E):=\int_E g(x)d\lambda_d$ for $E\subseteq U$, would have to be a Radon meausre, but I'm having trouble verifying outer regularity of this measure. Is it necessarily true that such a Borel measure is indeed Radon?

The definition for a Radon measure which I'm using is as follows:

Given a topological space $(X,\tau)$, a Borel measure $\mu:\mathcal{B}_X\rightarrow [0,\infty]$ is called a Radon measure if:

(1) $\mu(K)<\infty$ for all compact $K\subseteq X$.

(2) $\mu(U)=\sup \Big\{\mu(K): K\subseteq U, K \; \text{is compact} \Big\}$ for all $U\subseteq X$ open.

(3)$\mu(E)=\inf \Big\{\mu(V): E\subseteq V, V \; \text{is open} \Big\}$ for all $E\in \mathcal{B}_X$.

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  • $\begingroup$ Please define Radon measure for the community and tell us what is giving you trouble. $\endgroup$ – zhw. Sep 27 '18 at 14:43
  • $\begingroup$ I believe I was able to prove (1) and (2). However I was having trouble showing (3). $\endgroup$ – Keen-ameteur Sep 27 '18 at 16:47
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Yes, your $\mu$ is outer regular.

Lemma: Assume $f: U\to [0,\infty)$ is continuous. If $E\subset U$ is Borel and has compact closure in $U,$ then given $\epsilon>0,$ there exists $V$ open in $U$ such that $E\subset V$ and

$$\int_{V}f\,d\lambda <\int_{E}f\,d\lambda+\epsilon.$$

Proof: There is an open $W$ in $U$ such that $E\subset W,$ and $W$ lies in a compact subset of $U.$ By the outer regularity of $\lambda,$ there exist open subsets $V_k$ of $U$ such that $V_1\supset V_2 \supset \cdots \supset E$ and $\lambda(V_k\setminus E)<1/k.$ Note that continuity implies $f$ is bounded by some $M$ on $W\cap V_1.$ Thus

$$\int_{W\cap V_k}f\,d\lambda = \int_{E}f\,d\lambda+\int_{W\cap V_k \setminus E}f\,d\lambda $$ $$\le \int_{E}f\,d\lambda + M\cdot \lambda( W\cap V_k \setminus E) \le \int_{E}f\,d\lambda+\frac{M}{k}.$$

The last term will be $<\epsilon$ for large $k,$ giving the lemma.

To prove the outer regularity of $\mu,$ let $E\subset U$ be Borel. Let $\epsilon>0.$ Then we can write $E=\cup E_m,$ where the $E_m$ are pairwise disjoint, and each $E_m$ lies in a compact subset of $U.$ Use the lemma to see that there are open subsets $V_m$ of $U,$ with $V_m \supset E_m,$ such that

$$\int_{V_m}f\,d\lambda \le \int_{E_m}f\,d\lambda +\frac{\epsilon}{2^m}.$$

Then

$$\int_{\cup V_m}f\,d\lambda \le \sum_{m=1}^{\infty} \int_{V_m}f\,d\lambda$$ $$ \le \sum_{m=1}^{\infty} \left (\int_{E_m}f\,d\lambda +\frac{\epsilon}{2^m}\right) = \int_{E}f\,d\lambda +\epsilon.$$

Since $E\subset \cup V_m,$ we have shown the outer regulariy of $\mu.$

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  • $\begingroup$ Two follow up questions if you would agree: (1) Is there a need for the $E_m$-s to be disjoint? (2)Do you think this arguement would also work for a locally integrable function $g$? $\endgroup$ – Keen-ameteur Sep 28 '18 at 18:27
  • $\begingroup$ I need the $E_m$ to be disjoint for the last equality. (2) Yes I think locally integrable is enough; the argument would need to be changed only in the proof of the lemma $\endgroup$ – zhw. Sep 28 '18 at 19:16
  • $\begingroup$ Okay, thank you for your thorough answer. $\endgroup$ – Keen-ameteur Sep 28 '18 at 19:47
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The following is an answer to the original question which referred to the induced measure on all of $\mathbb{R}^d$

No, $\mu$ need not be outer regular. Let $d=1$, $U=\{r\in\mathbb{R}\mid r>0\}$ and $g:U\to\mathbb{R}$ be given by $g(x)=1/x$. Clearly, $\mu(\{0\})=0$, but every open set containing $0$ includes a set of the form $[0,\epsilon]$ with $\epsilon>0$ and $$\int_{[0,\epsilon]} g~\mathrm d\lambda=\infty.$$

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  • $\begingroup$ Given the clarfication I've written above (we assign measure to Borel subsets of $U$), is it still not true? $\endgroup$ – Keen-ameteur Sep 27 '18 at 16:44
  • $\begingroup$ $\mu$ is only defined on Borel subsets of $U$ which does not include $\{0\}.$ $\endgroup$ – zhw. Sep 27 '18 at 17:22

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