Borel measurable function that preserves Lebesgue measure Let $m$ the Lebesgue measure on the reals and $\mathcal{B}(\mathbb{R})$ the Borel sigma algebra. 
A transformation that preserves the Lebesgue measure is a Borel mensurable function $T:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $B\in\mathcal{B}(\mathbb{R})$, $m(B)=m(T^{-1}(B))$.
Show that, if $f:\mathbb{R}\rightarrow\mathbb{R}$ is a Borel measurable function and $m(f^{-1}(a,b]))=b-a$, $\forall$$a<b$ then $f$ is a transformation that preserves the Lebesgue measure.
I can't figure out how to prove this. Appreciate any suggestions.
 A: First, notice that the function $A\in\mathcal{B}(\mathbb{R})\mapsto  m(f^{-1}(A))\in\mathbb{R}$ is a measure. This measure is called the pushforward measure, and denoted by $(f_*m)$. Thus, you have to show that $m=(f_*m)$.
This question is a corollary of the uniqueness of extension of $\sigma$-finite measures:

If $\mu_1$ and $\mu_2$ are two measures on a $\sigma$-algebra $\mathscr{S}$ of a set $S$, $\mathscr{B}$ is a set that generates $\mathscr{S}$, $\mu_1(B)=\mu_2(B)$ for every $B\in\mathscr{B}$ and $S$ can be written as $S=\bigcup_{n=1}^\infty B_n$, where each $B_n\in\mathscr{B}$ satisfies $\mu_1(B)=\mu_2(B)<\infty$, then $\mu_1=\mu_2$.

The condition that $S$ can be written as a countable union of sets in $\mathscr{B}$ with finite measure is called $\sigma$-finiteness on $\mathscr{B}$. You can find the proof for this result in this page, $\S 9$. Here, you use $S=\mathbb{R}$, $\mathscr{B}$ the colletion of sets $(a,b]$ for $a<b$, $\mathscr{S}=\mathcal{B}(\mathbb{R})$, $\mu_1=m$ and $\mu_2=(f_*m)$.
