Measurability of the inverse of a measurable function Let $m$ denote Lebesgue measure, and let $f:[0,1] \to [0,1]$ be a (Lebesgue) measurable and bijective function. In general, it is not true that $f^{-1}$ is measurable. However, suppose that we now have the condition that $\forall A \subset [0,1]$, $m(A) = 0 \Rightarrow m(f(A)) = 0$. Why does this condition guarantee the measurability of $f^{-1}$? 
 A: The analogous fact about Borel functions is true without additional hypotheses: if $f: [0,1] \to [0,1]$ is a Borel bijection then its inverse is also Borel.  This follows from the observation that images of Borel sets under injective Borel functions remain Borel (this is far from obvious, and takes some descriptive set theory to prove -- see, e.g., 15.A of Kechris' Classical Descriptive Set Theory).
[Caveat: Injectivity of the Borel function in the previous observation is essential!  You can generalize slightly to Borel functions which are countable-to-one, but it's simply false for general Borel functions.]
Armed with this observation, we can also understand the result for measurable bijections.  Suppose that $f:[0,1] \to [0,1]$ is a measurable bijection which sends measure $0$ sets to measure $0$ sets. Fix a measure $1$ Borel set $A \subseteq [0,1]$ such that the restriction $f|_A : A \to [0,1]$ is Borel (in other words, for all Borel $Y \subseteq [0,1]$, the set $(f|_A)^{-1}(Y) = f^{-1}(Y) \cap A$ is Borel, not just Lebesgue measurable).
Edit: see below for more information about building such a set $A$.
To show that $f^{-1}$ is measurable, it suffices to show that $f(B)$ is measurable for all Borel $B \subseteq [0,1]$.  We see
$$
f(B) = f(B \cap A) \cup f(B \setminus A) = f|_A(B \cap A) \cup f(B \setminus A).
$$
Now $f|_A(B \cap A)$ is the image of a Borel set under an injective Borel function, thus Borel.  Also, $f(B \setminus A)$ is the image of a measure $0$ set (as $A$ had full measure), and is thus measure $0$ by hypothesis.  That means we've written $f(B)$ as the union of a Borel set with a measure $0$ set, establishing its measurability.
It's easy to see how this argument falls apart if $f$ no longer sends measure $0$ sets to measure $0$ sets.  The behavior of $f$ on $[0,1] \setminus A$ can be horribly pathological, making the $f(B\setminus A)$ part of the above expression some nonmeasurable set.

For the sake of clarity, here's an explanation of why there's a measure $1$ Borel set $A \subseteq [0,1]$ such that $f|_A : A \to [0,1]$ is Borel.  Let $(O_n)$ be an enumeration of a countable base for the topology on $[0,1]$ (for example, intervals with rational endpoints).  For each $n$, let $A_n \subseteq [0,1]$ be a measure $1$ Borel set such that $A_n \cap f^{-1}(O_n)$ is Borel (these exist by standard tightness of measure arguments).  Then $A = \bigcap_n A_n$ is again a measure $1$ Borel set, which we claim works.
First, note that for each $m$, $A \cap f^{-1}(O_m) = (\bigcap_n A_n) \cap (A_m \cap f^{-1}(O_m))$, which is Borel.  This implies that $A \cap f^{-1}(B)$ is Borel for all $B$ in the $\sigma$-algebra generated by $\{O_n\}$, which means that $f|_A$ is Borel.
