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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 $f^{-1}$?

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Since f is a bijection, $(f^{-1})^{-1}(A)=f(A)$. So you need to take A open in [0,1]-domain, and show that f(A) is measurable in [0,1]-codomain. But if A is open, it is the countable union of disjoint open intervals... – gary Aug 6 '11 at 22:39
Sorry, I don't see where in my suggestion we're using the condition that f preserves sets of measure 0; let me think it thru; I think you can approximate your measurable setO by the union $\cup O_i$ of open intervals, so that $m(O-\cup O_i)$=$\varepsilon $. Then play around with $(f^{-1})^{-1}=f$ to send sets (the non-zero measure part) back-and-forth. – gary Aug 6 '11 at 23:07
up vote 15 down vote accepted

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.

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Well done, ccc. – gary Aug 7 '11 at 5:45
@gary: Thanks! I wonder whether there's a proof of this that avoids using significant descriptive set theory. – user83827 Aug 7 '11 at 15:06
@ccc: Thanks for the help! Sorry if this is dumb, but can you explain what it means to fix a Borel set $A$ s.t. $f|_A$ is Borel? If $f$ is a Lebesgue measurable function, then it will still be defined on Lebesgue subsets of $A$ for any $A$ won't it? I think I understand the overall argument that you made though. Also, would you happen to have a reference for the fact that the images of Borel sets under injective Borel functions are Borel? – user1736 Aug 8 '11 at 4:46
@user1736: Don't apologize for my imprecisely written answer! I've added some more details, which I hope answer your questions. – user83827 Aug 8 '11 at 15:30
@ccc: Hm, I think I get the new defintion that you're using now, but is it obvious that you can find such an $A$ that restricts $f$ in such a way that $f^{-1}(Y) \cap A$ is always Borel? – user1736 Aug 8 '11 at 20:21

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