In this question, one of the answers claimed that the function $f: \mathbb{R}^{2n} \to \mathbb{R}$ given by $f(x,y) = x-y$ pulls back Lebesgue null sets to null sets, that is, $f^{-1}(N)$ is a null set for any null set $N$ and mentions it can be proved using Fubini's theorem, but gives no proof of this.
Could someone elaborate on why this must be true?
Let $S \subseteq \mathbb{R}$ be null. $$\begin{align*} \lambda^2(S) &= \iint_{\mathbb{R}^2} \chi_{f^{-1}[S]} d\lambda^2 = \int_{\mathbb{R}} \int_{\mathbb{R}} [x - y \in S] dx\,dy \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} \chi_{S + y}(x) dx\,dy \\ &= \int_{\mathbb{R}} 0\,dy = 0 \end{align*}$$ where $[\cdot]$ is the Iverson bracket. Notice that $x - y \in S \iff x \in S + y$. Since Lebesgue measure is invariant under translations, $S + y$ is again null.
