# If $\int_X f\varphi\,\mathrm{d}\mu\leq a$, then $\int_X f\,\mathrm{d}\mu\leq a$

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?

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.

• 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? $X$? Sep 7, 2020 at 7:11
• @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. Sep 7, 2020 at 19:11
• @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? Sep 7, 2020 at 19:18
• @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. Sep 7, 2020 at 19:39
• @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. Sep 7, 2020 at 20:20

Hint:

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)

• 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 ... ? Sep 7, 2020 at 5:32
• yes, but also to emphasize the topology on $X$. Sep 7, 2020 at 14:41
• 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 $X$? Sep 7, 2020 at 16:15