# Is the integral of a continuous map always a Radon measure?

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 $$E\in \mathcal{B}_X$$.

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

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 $$\mu.$$

• 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 $g$? – Keen-ameteur Sep 28 '18 at 18:27
• 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 – zhw. Sep 28 '18 at 19:16
• Okay, thank you for your thorough answer. – Keen-ameteur Sep 28 '18 at 19:47

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 $$\int_{[0,\epsilon]} g~\mathrm d\lambda=\infty.$$

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