Corollary 2

— Let L be a subset of G stable for multiplication and whose complement is locally negligible for a Haar measure $\beta$. Then $L = G$.

For, $L^{-1}$ and $L \cap L^{-1}$ have locally $\beta$-negligible complements. Now, $L \cap L^{-1}$ is a subgroup, hence is open (Cor. 1) and therefore closed. Therefore $G - (L \cap L^{-1})$, which is open and locally $\beta$-negligible, is empty. Thus $G = L \cap L^{-1}$.