Haar measure is positive on nonempty open sets and finite on compact sets

Since $I\mathrm{\ne }0$, some real $h$ has $I(h)\mathrm{\ne }0$. From $h=h^{+}-h^{-}$ and positivity at least one nonnegative part $g$ has $I(g)>0$. For $0\mathrm{\ne }f\ge 0$ choose $g\le \sum c_j{L}_{x_j}f$. Positivity and invariance imply $0<I(g)\le (\sum c_j)I(f)$, so $I(f)>0$.

If a nonempty open $U$ had $\mu (U)=0$, fix $u\in U$. The translates $x{u}^{-1}U$ cover $G$. Any compact $K$ is covered by finitely many and hence has zero measure. Open inner regularity would give zero measure to every open set, and outer regularity would then give zero measure to every Borel set, contrary to nonzeroness. Thus every nonempty open set has positive measure.

Compact finiteness is part of the definition. A compact neighbourhood of any point contains an open neighbourhood of no larger measure, proving local finiteness. The empty set still has measure zero and the zero test function has integral zero, so neither is incorrectly included in the strict positivity assertions.