Lebesgue Measurable/Integrable Functions and Singular Linear Transformations

The following theorem is paraphrased from Folland's Real Analysis: Modern Techniques and their Applications.

Suppose $T \in GL(n, \mathbb{R})$. If $f$ is a Lebesgue measurable function on $\mathbb{R}^{n}$, so is $f \circ T$. If $f \geq 0$ or $f$ is integrable with respect to the $n$-dimensional Lebesgue meseaure, then $$\int f(x) \mathrm{d}x = |\det T| \int f \circ T(x) \mathrm{d}x.$$ If $E$ is Lebesgue measurable, then $T(E)$ is also Lebesgue measurable, and $m(T(E)) = |\det T|m(E)$.

One of the exercises asks how much of this remains true if $T$ is not invertible.

Now, assuming that $f \circ T$ is Lebesgue measurable, we do not necessarily have $\int f = |\det T| \int f \circ T$ because $\int f$ might be nonzero, whereas $|\det T| = 0$. However, how do we show that $f \circ T$ is or is not Lebesgue measurable?

I started by assuming that $E \subset \mathbb{R}$ is a Borel set. Then $(f \circ T)^{-1}(E) = T^{-1}(f^{-1}(E))$. Since $f$ is Lebesgue measurable, we know that $f^{-1}(E)$ is a Lebesgue measurable set, so $f^{-1}(E) = F \cup N$, where $F$ is a Borel set and $N$ is a Lebesgue-null set. Then $$T^{-1}(f^{-1}(E)) = T^{-1}(F \cup N) = T^{-1}(F) \cup T^{-1}(N).$$ Of course, since $F$ is Borel and $T$ is continuous, we know that $T^{-1}(F)$ is Borel. But what do we do with $T^{-1}(N)$?

I am also unsure on how to show that $T(E)$ is Lebesgue measurable whenever $E$ is Lebesgue measurable, but I believe this is true.

As for the claim that $m(T(E)) = |\det T| m(E)$, since $\det T = 0$, we would have to show that $m(T(E)) = 0$. Since $T$ is singular, $T(E)$ is a subset of some subspace of $\mathbb{R}^{n}$ that is not the whole space, so $T(E)$ is a subset of a null set. Since Lebesgue measure is complete, $m(T(E)) = 0$. (I still have to formalize this idea, but I think I'm on the right track here.)

"What do we do with $T^{-1}(N)$?" We use it to build a counterexample. Let $f:\mathbb R^2\to\mathbb R$ be a nice well-behaved function, say identically zero, except that on some proper linear subspace, say the $x$-axis, it's completely wild. So $f$ is Lebesgue measurable because its wildness is on a set of measure zero. But if $T$ is the projection of $\mathbb R^2$ onto the $x$-axis, then in $f\circ T$ the wildness permeates all of $\mathbb R^2$ and you can thus arrange for $f\circ T$ not to be Lebesgue measurable.