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A Haar measure is a Borel measure $\mu$ in a locally compact topological group $X$, such that $\mu$(U)>0 for every non empty Borel open set $U$, and $\mu(xE)$=$\mu(E)$ for every Borel set $E$.

A Borel measure is a measure $\mu$ defined on the class $S$ of all Borel sets and such that $\mu$(C)< $\infty$ for every C in $C$.

we shall denote by $C$ the class of all compact subsets of X ,by S the $\sigma$-ring generated by $C$.we shall call the sets of S the borel sets of X.

We want to show that the first property is equivalent to the assertion that $\mu$ is not identically zero.

If $\mu(U)$=0 for some non empty Borel open set $U$, and if $C$ is any compact set,then the class $\{xU\mid x \in C\}$ is an open covering of $C$. Since $C$ is compact, there exists a finite subset $\{x_1,\dots,x_n\}$ of $C$ such that $C \subset \bigcup\limits_{i=1}^{n}x_i U$, and the left invariance of $\mu$ implies that $$\mu(C)\leq\sum_{i=1}^{n}\mu(x_iU)=n\mu(U)=0. $$ Since the vanishing on the class $C$ of all compact sets implies its vanishing on the class $S$ of all Borel sets, we obtain the desired result.

My question is:

Why is the class $\{xU\mid x \in C\}$ an open covering of $C$? And why does the vanishing on the class $C$ of all compact sets implies its vanishing on the class $S$ of all Borel sets?

Thanks very much in advance.

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  • $\begingroup$ The assumption that $\mu(C)<\infty$ for all $C\in S$ is very strange--it actually implies that $X$ must be compact, rather than locally compact. $\endgroup$ – Eric Wofsey Dec 2 '15 at 21:53
  • $\begingroup$ Also, what is your definition of "Borel set"? From the fact that you are referring to "Borel open sets", I assume your definition is the $\sigma$-ring generated by compact sets rather than the $\sigma$-algebra generated by open sets? $\endgroup$ – Eric Wofsey Dec 2 '15 at 22:16
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Let me begin by clarifying a few things about the definition. First, I assume from your phrasing that you are defining the "Borel sets" to be the $\sigma$-ring generated by the compact subsets of $X$ (rather than the $\sigma$-algebra generated by the open subsets of $X$). Second, the definition of Haar measure should not say that $\mu(C)<\infty$ for all Borel sets $C$; it should only say that $\mu(C)<\infty$ for all compact sets $C$. If you assume $\mu(C)<\infty$ for all Borel sets, this actually implies $X$ must be compact (see this question).

Let me now address your questions. For the first question, it is true that $\{xU\mid x\in C\}$ covers $C$ if $U$ contains the identity $1\in G$, since then $x\in xU$ for each $x\in C$. But the $U$ we were given might not contain the identity, and in that case there is no reason to expect $\{xU\mid x\in C\}$ to cover $C$. To fix this, just let $y\in U$ be any point, and replace $U$ with $y^{-1}U$.

For the second question, simply note that the collection of Borel sets $E$ such that $\mu(E)=0$ is a $\sigma$-ring, so if it contains all compact sets, then by definition it contains all Borel sets.

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  • $\begingroup$ Additional note: If we define the Borel sets as usual, then vanishing on all compact sets implies vanishing on all open sets, since usually it is required that the Haar measure is Radon, which in particular implies $\mu (U)=\sup_{K\subset U}\mu (K) $ (with K compact) for each open U and $\mu (E)=\inf_{U\supset E}\mu (U) $ (with U open) for arbitrary Borel sets. $\endgroup$ – PhoemueX Dec 3 '15 at 7:43
  • $\begingroup$ for the second question,in proving that the collection of Borel sets $E$ such that $\mu(E)=0$ is a $\sigma$-ring in the part of proving that it is closed under difference if we suppose that E is a borel set s.t. $\mu$(E)=0 and F is a borel set s.t. $\mu$(F)=0 then E-F is a borel set is ok for me but how to prove $\mu$(E-F)=0. for the first question,it is from halmos's book "measure theory"p.252 i still have the same question and how he use also that U is open in proving that the class $\{xU\mid x \in C\}$ is an ""open"" covering of $C$. $\endgroup$ – aya_haitham Dec 7 '15 at 20:16
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    $\begingroup$ Since $E-F\subseteq E$, $\mu(E-F)\leq \mu(E)=0$. If $U$ is open, then $xU$ is open for any $x$ because multiplication by $x$ is a homeomorphism (it is continuous, as is its inverse, multiplication by $x^{-1}$). $\endgroup$ – Eric Wofsey Dec 7 '15 at 21:10

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