regularity of $d\mu=u dx$

Question

Let $G$ be a abelian, locally compact group with a Haar measure $dx$ on $G$. We know that every Haar measure is inner regular, ie... for any open subset $U$ of $G$, then $dx(U)=\sup\{dx(F):\;F\;\;\text{is compact and is a subset of}\;\;U\}$.Here we denote $dx(F)$ is the measure of $F$ respect to $dx$.

Now we are given a non-negative function $u:G\to\mathbb R$, an open subset $U$ of $G$ such that $u\in L^1_{loc}(G)$. We assume that, there is a constant $c>0$ so that $\int_F u(y)dy\leq c<\infty$ for any $F$ is compact and is a subset of $U$.

Is it true that $\int\limits_Uu(y)dy=\sup\limits_F\;\int\limits_F u(y)dy$, where supremum is taken over all compact $F$ which is a subset in $U$? Could you give me some ideas to prove if it is true.

Answer 1

The following steps lead to a solution if $U$ has compact closure:

(1) Firstly, prove that $\int_F u(y)dy\leq \int_U u(y)dy$ for all compact $F\subseteq U$. (Hint: $u:G\to\mathbb{R}$ is $\textit{nonnegative}$.)

(2) Secondly, we wish to prove that for each $\epsilon>0$, there is a compact $F\subseteq U$ such that $\int_U u(y)dy - \int_F u(y)dy<\epsilon$. Note that $\int_U u(y)dy -\int_F u(y)dy = \int_{U\setminus F} u(y)dy$.

(3) Since $u\in L_{\text{loc}}^1(G)$, $u\in L^1(\overline{U})$ where $\overline{U}$ denotes the closure of $U$. (We have used the assumption that $U$ has compact closure.)

(4) Prove the following general fact in measure theory: if $(X,\mu)$ is a measure space and if $f\in L^1(X)$, then to each $\epsilon>0$, there exists a $\delta>0$ such that whenever $\mu(E)<\delta$, then $\int_{E} \left|f\right|<\epsilon$.

(5) If $\epsilon>0$, apply the inner regularity of the Haar measure $dx$ to conclude the existence of a compact $F\subseteq U$ such that the measure of $U\setminus F$ is $<\delta$ where $\delta$ is as in (4) (with $u$ in place of $f$ and $\overline{U}$ in place of $X$). Conclude the result from (2).

I hope this helps!

Answer 2

If we know that $U$ is $\sigma$-compact (i.e., that $U$ is a countable union of compact sets), then there is an easy proof.

(1) Since $U$ is $\sigma$-compact, we can write $U=\bigcup_{n=1}^{\infty} F_n$ where each $F_n$ is a compact subset of $U$ and $F_n\subseteq F_{n+1}$ for all positive integers $n$. Let $u_n=u\chi_{F_n}$ (let us recall that $\chi_{F_n}$ is the characteristic function of $F_n$) and observe that $0\leq u_1\leq u_2\leq\cdots\leq u$ and that $u_n\to u$ pointwise on $U$. Use the monotone convergence theorem to deduce that $\int_{U} u(y)dy=\lim_{n\to\infty} \int_{F_n} u(y)dy$. Therefore, your result is true in the case when $U$ is $\sigma$-compact. (Note that this is the case in most natural contexts.)

(2) Of course, $U$ is a locally compact Hausdorff space in any case. Prove that if we know, in addition, that $U$ is $\textit{second countable}$, then it follows that $U$ is $\sigma$-compact. Therefore, your result is true if $G$ is also assumed to be second countable.