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By a Haar measure on a locall compact group (Hausdorff) we mean a positive measure $\mu$ (contains the borel set's) such that

  • The measure $\mu$ is left invariant

  • The measure μ is finite on every compact set

  • Is $\mu$-regular (i.e. outer and inner regular)

1) It can be shown as a consequence of the above properties that $\mu(U) > 0$ for every non-empty open subset $U$.

Why $\mu(U)>0$ if $U$ is open?

Thank you all.

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1 Answer 1

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If $\mu \neq 0$, then by the inner regularity there is a compact $K \subset G$ with $\mu(K) > 0$. Let $u\in U$. Then

$$\mathscr{C} = \{ xu^{-1}U : x \in K\}$$

is an open cover of $K$. By compactness, there is a finite subset $F\subset K$ such that

$$\{ xu^{-1}U : x \in F\}$$

covers $K$. Then

$$0 < \mu(K) \leqslant \mu\left(\bigcup_{x\in F} xu^{-1}U\right) \leqslant \sum_{x\in F} \mu(xu^{-1}U) = \sum_{x\in F} \mu(U) = \mu(U)\cdot \operatorname{card} F$$

by monotonicity and subadditivity of measures, and left-invariance of Haar measures.

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    $\begingroup$ You could simplify this slightly by just using the open cover $\{xU : x \in G\}$. $\endgroup$ Dec 2, 2014 at 22:27

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