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Can we have a measurable function $f$, whose inverse is not measurable?

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If you're considering only Lebesgue measure, this is Problem 2.2.19 in Problems in Mathematical Analysis III: Integration by Kaczor, Kaczor, and Nowak. The book also contains worked solutions for all of the problems. The answer is yes. – Jonas Meyer Oct 11 '10 at 7:23
up vote 3 down vote accepted

I think I have an answer that works now.

Obviously the problem is trivial if we allow $f$ to be any function between measure spaces (just take the domain to have its powerset as $\sigma$-algebra and/or the trivial $\sigma$-algebra on the codomain), so this must be a question about real functions and the Lebesgue measure.

Let $X$ be the Cantor set, $Y$ a nonmeasureable set (and hence uncountable) contained in $[0,1]$ and $g: X \to Y$ be any bijective function. Then define $$f(x) = \begin{cases} g(x) & \text{if } x \in X \\ e^x + 2 & \text{otherwise.} \end{cases}$$

Now let $U$ be any measurable set and let us show that $f^{-1}(U)$ is measurable. If $U \cap Y = \emptyset$ then $f^{-1}(U) \subset \mathbb{R} \setminus X$ on which $f$ is continuous. In particular $f^{-1}(U)$ must be measurable. If $U \cap Y \neq \emptyset$ then we split into two more cases. First consider the possibility that $U \cap (2,\infty) = \emptyset$. Then $f^{-1}(U) \subset X$, i.e. it is a null set and hence Lebesgue measurable. And if $U \cap (2,\infty) \neq \emptyset$ then by a similar argument we see that $f^{-1}(U)$ is the disjoint union of a null set contained in $X$ and a measurable subset of $\mathbb{R}\setminus X$.

Thus $f$ is measurable, but its inverse is obviously not measurable since $(f^{-1})^{-1}(X) = f(X) = Y$.

Now if we also require $f$ to be bijective rather than just injective, then I don't know the answer. It seems much harder to construct an example in that case.

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My previous answer is wrong.

Take $f(x) = x$ ($f: \mathbb{R} \to \mathbb{R}$) where the domain has the Borel $\sigma$-algebra and the range the trivial $\sigma$-algebra. Then $f$ is clearly measurable but $f$ maps for example $[0,1]$ to $[0,1]$ which is not measurable.

Or do you want both $\sigma$-algebra's to be Borel? In that case I should modify my first answer (which was wrong because not every subset of a measure zero set is Borel measurable).

Another idea would be the following (based on the one below me):

Let $g:[0,1] \to \mathbb{R}$ be the Cantor function. Extend it to all of $\mathbb{R}$ by defining $g(x) = 0$ for $x < 0$ and $g(x) = 1$. Now define $h(x) = x + g(x)$ (this is standard), this function has a lot of nice properties like: $h$ is a bijection and measure of $h(C) = 1$ where $C$ is the Cantor set.

Now we know that a set of positive measure contains a non-measurable subset. So let $A$ be an non-measurable subset of $h(C)$. So $M = h^{-1}(A)$ is a subset of the Cantor set and has thus zero Lebesgue measure.

So define $u := h^{-1}$. This function is measurable (Lebesgue/Borel). But $u^{-1}(M) = A$ is non-measurable.

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While it is true that not every subset of a Borel set of measure zero is a Borel set, it is indeed true that every subset of a null set is Lebesgue measurable - that's part of the reason we go to the trouble of showing that the Lebesgue measure is complete. And the reason why it's possible to construct an example such as the one in my answer (although it's rather contrived and honestly not that good). – kahen Oct 11 '10 at 18:47
Yes, I modified it a bit and based my answer on yours. If it is correct, you deserve the credit ;-). – Jonas Teuwen Oct 11 '10 at 18:49

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