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.
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.
