Why is $\Delta(f)(s,t)= f(st)$ well-defined?

Question

Let $G$ be a locally compact group with Haar measure $\lambda$. The locally compact group $G\times G$ then carries the Haar measure $\lambda\times \lambda$. One can then define the map $$\Delta: L^\infty(G)\to L^\infty(G\times G, \lambda\times \lambda)$$ by $$\Delta(f)(s,t) = f(st).$$ Why is this well-defined? I.e. if $f= 0$ almost everywhere, why is $\Delta(f)= 0$ almost everywhere? This seems to be well-known, but not so obvious to me: Why is $$\{(s,t): f(st)\ne 0\}$$ a null-set, if $\{f\ne 0\}$ is a null-set?

Answer 1

Denote $\cdot : G\times G \to G$ to be the group operation. We will think about it as function from product to $G$ most of the times. For $f \in L^\infty(G, \lambda)$, the function $\Delta f$ is composition $f \circ \cdot$. You are essentially asking why $$\lambda(f^{-1}(\mathbb{R} \setminus \{0\})) = 0\quad\implies\quad \lambda \times \lambda ((f \circ \cdot)^{-1} (\mathbb{R} \setminus \{0\})) = 0.$$

This can be proven using the defining property of product measure, and invariance of Haar measure with respect to $\cdot$. Product measure is one satisfying $$ \lambda \times \lambda(A) = \int_G \lambda(A^g) \lambda(dg),$$ where $A \subseteq G\times G$ and $A^g = \{h \in G: (g, h) \in A\}$. With that, we can write for any $\lambda$-null set $Z \subseteq G$ that $$ \lambda \times \lambda(\cdot^{-1} Z) = \int_G \lambda((\cdot^{-1} Z)^g) \lambda(dg) = \int_G \lambda(g^{-1}Z)) \lambda(dg), $$ because $(\cdot^{-1} Z)^g = \{h: g \cdot h \in Z\}$. Now using invariance, function under the integral is always $\lambda(Z) = 0$, and therefore integral is zero as well.

The result now follows by taking $Z = \{g: f(g) \neq 0\}$, because $\cdot^{-1} Z$ is then precisely $\{(g, h): f(gh) \neq 0\}$.