Lebesgue-measurable function problem

Question

I am trying to solve this exercise:

Let $\phi: \mathbb R \to \mathbb R$ be a measurable function (Lebesgue-measurable) and $f:\mathbb R^2 \to \mathbb R$ defined as $f(x,y)=xy-\phi(y)$. Show the following:

a) $f$ is measurable

b) If $E \subset \mathbb R$ measurable, then $f^{-1}(E)$ is measurable.

This is what I could do:

a) If I define $h:\mathbb R^2 \to \mathbb R$ as the projection $h(x,y)=y$, then $h$ is continuous and $\phi(y)=\phi \circ h(x,y)$, if I could prove that $\phi \circ h$ is measurable, then, since $g(x,y)=xy$ is continuous and $f=g+ \phi \circ h$ is sum of measurable functions, $f$ is measurable.

I couldn't show that $\phi \circ h$ is measurable, so I am not sure if my approach is the appropiate one.

b) If $E$ is measurable, then $E$ can be written as $E=H \setminus Z$ with $H$ a $G_{\delta}$ set and $m(Z)=0$. The preimage is $f^{-1}(H \setminus Z)=f^{-1}(H) \setminus f^{-1}(Z)$. Since $H$ is a borel set, then it is easy to see that $f^{-1}(H)$ is measurable.

I am having some difficulty trying to show that $f^{-1}(Z)$ is measurable. Since $m(Z)=0$, there exists $G$ a $G_{\delta}$ set with $Z \subset G$ and $m(G)=m(Z)=0$. The set $G$ can be written as $G=\bigcap_{k \in \mathbb N} G_k$ with $(G_k)_{k \in \mathbb N}$ decreasing, $m(G_k)<\infty$. We have $f^{-1}(Z) \subset f^{-1}(G) \subset f^{-1}(G_k)$ for all $k \in \mathbb N$. I didn't know what to do next.

Any suggestions would be greatly appreciated. Thanks in advance.

Answer 1

Hint for (a): If $B$ is a Borel subset of ${\Bbb R}$, then $(\phi\circ h)^{-1}(B) = {\Bbb R}\times\phi^{-1}(B)$, and $\phi^{-1}(B)$ is a (Lebesgue) measurable set because $\phi$ is assumed to be Lebesgue measurable.

Answer 2

The trick is to show that the inverse image of a set of a set of measure zero is measure zero, (and hence measurable). Let $Z \subset \mathbb R$ be measure zero. Let $\lambda_n$ denote Lebesgue measure on $\mathbb R^n$. $$ m = \lambda_2(\{(x,y):xy-\phi(y) \in Z\}) = \int \lambda_1(\{x:xy-\phi(y) \in Z\}) \, d\lambda_1(y) .$$ Unless $y = 0$ (and $\{0\}$ is a set of measure zero, so can be disregarded) $$ \{x:xy-\phi(y) \in Z\} = \frac 1 y (Z + \phi(y)) ,$$ and so $$ \lambda_1(\{x:xy-\phi(y) \in Z\}) = \frac1{|y|} \lambda_1(Z) = 0.$$ Hence $m = 0$.