Lebesgue measure of set not equal to that of its preimage

Question

Let $m$ be the Lebesgue measure on $\mathbb{R}$ and let $f:\mathbb{R}\to[0,\infty)$ be a Lebesgue integrable function. Then, does there exist a set $E\subseteq[0,\infty)$ such that $m(E)\neq m(f^{-1}(E))$?

I would take $f$ to be $x^2$ whence any nontrivial interval would satisfy the above property. But, $x^2$ is not integrable on the real line. How do we prove in the general case? Any hints? Thanks beforehand.

Answer 1

If $f\ge0$ is integrable and $E_k=[2^k, 2^{k+1})$ then $$\sum_{k\in\Bbb Z}2^km(f^{-1}(E_k))<\infty.$$

Details:

Lemma. If $\lambda\in\Bbb R$ then $\sum_{k\in\Bbb Z, 2^k<\lambda}2^k\le 2\lambda.$

Proof: Exercise.

Now say $f\ge0$ and $f$ is integrable. Define $E_k$ as above, let $A_k=f^{-1}(E_k)$, and let $$g=\sum_{k\in\Bbb Z}2^k\chi_{A_k}.$$The lemma shows that $g\le 2f$, so $$\sum 2^km(A_k)=\int g<\infty.$$

Answer 2

Take $f(x)= \frac{1}{1+x^2}$ and $E=[2, 3]$.

Then $m(E)=1$, $f^{-1}(E)= \emptyset$ and $m (f^{-1}(E))=0$.