11

Let $f :\mathbb R \to \mathbb R$ be a bijective Lebesgue measurable function , then is $f^{-1}:\mathbb R \to \mathbb R$ Lebesgue measurable ? I don't think this is true but I can't find any counterexample , or any way through . Please help . Thanks in advance

NOTE : Any counter example using non Lebesgue measurable set is fine with me

  • Your thought is correct, but finding such counterexample involves finding a non Lesbesgue measurable set, which is not trivial. – Simon Marynissen Mar 23 '17 at 18:14
  • 2
    @stochasticboy321 Some problems with what you propose. You say that $Y\subset [0,1]$ is nonmeasurable. Why does that imply $Y$ has the cardinality of the Cantor set? (It doesn't, but that's fixable). More important is that we are looking for a bijection from $\mathbb R$ to $\mathbb R.$ Your function isn't, although it is injective. – zhw. Mar 23 '17 at 19:22
  • @zhw. Indeed, you're right. I seem to have had a brainfart. Deleted. – stochasticboy321 Mar 23 '17 at 22:21
  • @stochasticboy321 Ah, no need to delete it, but whatever ... It is the same place I started. See my answer below. – zhw. Mar 23 '17 at 22:54

1 Answers1

4

The interval $(0,1/4]$ contains a nonmeasurable subset (meaning a non Lebesgue measurable subset). The union of this set with $(1/4,1/2]$ is then a nonmeasurable subset $E$ of $(0,1/2]$ that has cardinality $c.$

Let $K_1,K_2$ be two disjoint Cantor sets, each a subset of $(0,1),$ having measure $0.$ Each of these has cardinality $c.$ Note that $(0,1)\setminus E$ also has cardinality $c.$ Thus there exists a bijection $f: K_1\cup K_2 \to (0,1)$ with $f(K_1) = E,$ $f(K_2)=(0,1)\setminus E.$

We can extend $f$ now to $K_1\cup K_2 \cup\mathbb Z$ by defining $f(n) = n$ for each $n\in \mathbb Z.$ Note that so far, we have only defined $f$ on a set (a Borel set in fact) of measure $0.$ This of course cannot harm Lebesgue measurability in any way as we move towards defining $f$ on all of $\mathbb R$ (although it could muck up Borel measurability).

Onwards: $K_1\cup K_2 \cup \mathbb Z$ is a closed subset of $\mathbb R.$ Hence its complement is the pairwise disjoint union of countably many open intervals $I_1,I_2, \dots.$ The idea is to map each of $I_1,I_3,I_5, \dots$ bijectively onto $(-1,0),(-2,-1),(-3,-2),\dots$ respectively. This can be done with homeomorphisms. Same with $I_2,I_4,I_6, \dots$ and the intervals $(1,2),(2,3),(3,4),\dots.$ Choose whatever homeomorphisms you like for this.

We now have $f$ defined as a bijection of $\mathbb R$ to $\mathbb R.$ And it is Lebesgue measurable, as you should verify.

Is $f^{-1}$ Lebesgue measurable? If it were, then $(f^{-1})^{-1}= f$ would take Borel sets to Lebesgue measurable sets. But as we've seen, $f(K_1) = E,$ and $E$ is not Lebesgue measurable.

zhw.
  • 105,693
  • 1
    Why would $(f^{-1})^{-1}$ take Borel sets to Borel sets? From Lebesgue measurability of $f^{-1}$, you only get that $(f^{-1})^{-1}$ takes Borel sets to Lebesgue measurable sets (am I missing something?). Of course, this suffices for what you want. Anyway, this also raises the question if the claim is valid when "Lebesgue measurable" is replaced by "Borel measurable" :) EDIT: Ok, as claimed here (http://math.stackexchange.com/questions/56022/measurability-of-the-inverse-of-a-measurable-function), the claim is true for Borel measurable functions. – PhoemueX Mar 24 '17 at 13:24
  • @PhoemueX Thank you, you are right about that bit at the end. I don't know what I was thinking. I'll edit. And thanks for link on the Borel measurable case. – zhw. Mar 24 '17 at 15:46