Suppose that $X$ is locally compact Hausdorff space. Let $f:X\to (0,\infty)$ continuous, and let $\varphi\in C_c(X)$ with $0\leq \varphi \leq 1$. Let $a>0$, and $\mu$ a Radon measure on $X$. I read in a proof, that if once proven that $\int_X f\varphi\,\mathrm{d}\mu\leq a$, then $\int_X f\,\mathrm{d}\mu\leq a$. The author said, the conclusion follows from the following lemma, which I can't see how it has been used

Lemma: Let $X$ Hausdorff space, $\mu$ Radon measure on $X$, and $\mathcal{K}$ the collection of compact subsets of $X$. If $f:X\to [0,\infty]$ is a Borel measurable function, then $$ \int_{X}f\,\mathrm{d}\mu=\sup_{K\in \mathcal{K}}\int_{K}f\,\mathrm{d}\mu. $$

Question: If $\int_X f\varphi\,\mathrm{d}\mu\leq a$, how to show that $\int_X f\,\mathrm{d}\mu\leq a$ by lemma above?


2 Answers 2


Using Urysohn's lemma, for each compact set, you can construct $\phi \in C_c(X)$ s.t. $0 \leq \phi \leq 1$ and $\phi|_K = 1$, then $\int_K f \ d\mu \leq \int f \phi \ d\mu$. Then if you prove that $\int f \phi \ d\mu \leq a$ for each $\phi \in C_c(X)$, then by the lemma you cite, $\int f \ d\mu= \sup_{K \in \mathcal K} \int_K f \ d\mu \leq a$.

You definitely need that inequality to hold for all $\phi \in C_c(X)$, otherwise just consider $\phi =0 $ and it holds trivially for any positive $a$.

Edit: More details on constructing the $\phi$.

First a statement of Urysohn's lemma for LCH spaces:

If $X$ is a locally compact Hausdorff (LCH) space and if $K, F \subseteq X$ are disjoint sets s.t. $K$ is compact and $F$ is closed, then there is a continuous function $\phi :X \to [0,1]$ s.t. $\phi|_K = 1$ and $\phi|_F = 0$.

Now our goal is that given any compact set $K$, we want to construct a $\phi \in C_c(X)$ s.t. $\phi|_K = 1$. Well we definitely want to use Urysohn's lemma, but what is our set $F$? Note it can't be $X^C = \emptyset$ as then we can't prove that the support is compact.

The following claim should suffice:

Let $X$ a LCH space. Let $K \subseteq X$ compact. Then there exists an open set $U$ s.t. $K \subseteq U$ and $\overline{U}$ is compact.

Prove this claim and use it to finish the proof.

  • $\begingroup$ In Ursysohn's lemma, it talks about a compact set $K$ and an open set $U$, satisfying that $K\subseteq U$, then there exists $f\in C_c(X)$ such that $0\leq f\leq 1$, $f|_K\equiv 1$ and $\textrm{supp} f\subseteq U$. What should $U$ be in this problem? $X$? $\endgroup$
    – James2020
    Sep 7, 2020 at 7:11
  • $\begingroup$ @James2020: You may have to use the fact that if $f$ is a nonnegative lower semicontinuous function on $X$, then $f=\sup\{\phi\in\mathcal{C}_{00}(X): 0\leq \phi\leq f\}$ along with the fact that ever open set $G$, or rather $\mathbb{1}_G$, is lower semicontinuous. $\endgroup$ Sep 7, 2020 at 19:11
  • $\begingroup$ @OliverDiaz I am not familiar with this one ... Wouldn't it possible to use only two lemmas: Urysohn's lemma and the lemma I stated in the post? $\endgroup$
    – James2020
    Sep 7, 2020 at 19:18
  • $\begingroup$ @James2020 I can add more detail. Though, you should try to figure it out yourself. I'll add some hints to my post, ask if you can't figure it out. $\endgroup$ Sep 7, 2020 at 19:39
  • 1
    $\begingroup$ @KeeferRowan: James's version of Urysohn's lemma includes the conclusion that $\phi \in C_c(X)$, so the support of $\phi$ is automatically compact. The lemma says only that the support of $\phi$ is contained in $U$; there is nothing stopping it from being much smaller. $\endgroup$ Sep 7, 2020 at 20:20


Let $\mathcal{K}$ denote the collection of all compact sets in $X$.

  • Using Urysohn's lemma one can show that $\int_K f\,d\mu\leq a$ for all compact.

  • On the other hand, $\nu(g)=\int g\,f\,d\mu$ defines a Radon measure (regular Borel measure) on $X$. By inner regularity $\mu(A)=\sup\{\nu(K): K\in\mathcal{K},\,K\subset A\}$ for all measurable sets $A$. (This is the Riesz-Markov representation theorem)

  • $\begingroup$ If I may ask, why do you say ".. compact sets in $X$" (and not "... compact subsets of $X$")? Is this a short-hand for saying that these sets are subsets of $X$, and that they are compact in X? Or ... ? $\endgroup$
    – James2020
    Sep 7, 2020 at 5:32
  • $\begingroup$ yes, but also to emphasize the topology on $X$. $\endgroup$ Sep 7, 2020 at 14:41
  • $\begingroup$ OK, glad to know it. You wrote in your first point about Urysohn's Lemma. The version I have in the book is: "For given sets $K\subseteq U$ in a locally compact Hausdorff space $X$, where $K$ is compact, and $U$ is open, there is $f\in C_c(X)$ such that $0\leq f\leq 1$, $f\equiv 1$ on $K$ and $\textrm{supp} f\subseteq U$." The question is, what is $U$ is this situation, when trying to prove the claim I wrote in the post? Should $U$ be $X$? $\endgroup$
    – James2020
    Sep 7, 2020 at 16:15

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