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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.

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Dear H D Hung, I notice that you have asked $9$ questions but that you have not accepted a single answer. Do you know how to accept answers to your questions? (If not, there is a green "tick" mark next to every answer and in order to accept an answer, you need to click on the green tick mark.) – Amitesh Datta Jun 23 '11 at 11:50
Dear Amitesh Datta, Thank you, I will take a look at it in the next post. – Jie Fan Jun 24 '11 at 1:21
up vote 4 down vote accepted

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!

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Yes, it's really help. Thanks~ – Jie Fan Jun 24 '11 at 1:21
Dear H D Hung, you are welcome! If you are satisfied with my answer, you can accept it by clicking the green "tick" mark to the left of it. – Amitesh Datta Jun 24 '11 at 1:36

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.

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The result is true for any locally compact space. Indeed, the functional $f \mapsto \int fu\,dx$ is positive and linear on the space of continuous functions with compact support and the associated (inner) measure is in particular inner regular by the Riesz representation theorem, see e.g. Pedersen, Analysis now, Theorem 6.3.4, p.237. Of course, that we're working on a locally compact space is the essential point here, the (abelian) group structure is completely immaterial. Also, the rather strange condition $\int_C u \lt c$ for all compact $C$ is of absolutely no use either. – t.b. Jun 23 '11 at 14:46
Dear Theo, you are absolutely correct; I completely forgot about this result! Thanks for pointing this out! I was somehow influenced by $\sigma$-compactness because of the result: every positive Borel measure that is finite on compact sets on a locally compact, Hausdorff space in which every open set is $\sigma$-compact is regular. (And a clean proof of this result uses Riesz.) I should have recognized the linear functional you mentioned especially since I was learning about distributions lately (and if $G=\mathbb{R}^n$, I suppose $u$ would be a distribution). – Amitesh Datta Jun 24 '11 at 0:52

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