Suppose $G$ is a Lie group and $A\subseteq G$. Suppose moreover that $(U_i,\varphi_i)$ is a cover of $G$ by charts. I know of two definitions for whether $A$ is a null set or not:
- $A$ has Haar measure zero.
- $\varphi_i(U_i\cap A)$ has Lebesgue measure $0$ for every $i$.
For the sake of this question, let's say $A$ is Haar null if it satisfies the first definition and Lebesgue null if it satisfies the second. I am sure these two definitions are equivalent, but I'm not sure how to prove it.
Clearly, $A$ is Haar null if and only if $A\cap U_i$ is Haar null for every $i$, so we can assume $A\subseteq U$ for some $U$ which admits a diffeomorphism $\varphi: U\to \Bbb{R}^n$, and ask whether $A$ is Haar null if and only if $\varphi(A)$ is Lebesgue null.
I'm pretty sure I know how to prove that if $A$ is Lebesgue null then it's also Haar null. For some differential form $\omega$ on $G$, the Haar measure of $A$ is $$\int_G \chi_A \omega=\int_{\Bbb{R}^n}(\varphi^{-1})^*\chi_A \omega=\int_{\Bbb{R}^n}(\chi_A\circ\varphi^{-1})(\varphi^{-1})^* \omega=\int_{\Bbb{R}^n}\chi_{\varphi(A)}(\varphi^{-1})^* \omega=0.$$ (Since $\varphi(A)$ is Lebesgue null).
So I have two questions:
- Is my proof correct?
- How do you prove every Haar null set is Lebesgue null?
I am sure this is trivial if you are familiar with these things. Perhaps you can say something about the the differential form $(\varphi^{-1})^* \omega$ in relation to $\wedge dx_1\cdots\wedge dx_n$?