These are some questions that I couldn't answer after a class I'm teaching. I would very much appreciate help.
Let $\Omega=[0,1]^N$ and suppose that $f,g\colon \Omega \to [0,1]^N$ are two functions that generate exactly the same $\sigma$-algebra on the domain $\Omega$. That is, for every Borel set $U\subseteq [0,1]^N$ there is a Borel set $W\subseteq [0,1]^N$ such that $f^{-1}(U)= g^{-1}(W)$; and for every Borel set $U\subseteq [0,1]^N$ there is a Borel set $W\subseteq [0,1]^N$ such that $g^{-1}(U)= f^{-1}(W)$.
What conditions guarentee that $f(g^{-1}(X))$ is always Lebesgue null whenever $X\subseteq [0,1]^N$ is a Borel set of Lebesgue measure zero?
I'm looking for a counterexample when $f,g$ are not measurable. How about if $f,g$ are measurable or continuous? What if $f,g$ are homeomorphisms? Smooth functions that preserve the boundary?
Measurability is not enough. Let $C$ be a closed, uncountable subset of of measure zero. By the Borel isomorphism-theorem, there is a Borel isomorphism $j:[0,1]^N\to C$. Of course, we can view $j$ also as a measurable function $j:[0,1]^N\to [0,1]^N$.
Let $f:\Omega\to[0,1]^N$ be some bijection and $g=j\circ f$. Clearly, $\sigma(f)=\sigma(g)$. Take any measurable set $B\subseteq [0,1]^N$ of positive measure and let $X=j(B)$, a measurable set since $j$ is a Borel isomorphism. Also, $X\subseteq C$, so $X$ has measure zero. Now,
$$f\big(g^{-1}(X)\big)=f\Big((j\circ f)^{-1}(X)\Big)=f\Big(f^{-1}\circ j^{-1}(X)\Big)=j^{-1}(X)=j^{-1}\big(j(B)\big)=B.$$