Function $f:[0,1] \rightarrow \Bbb R$ such that $\forall c \in \Bbb R: f^{-1}(c)$ Lebesgue Measurable, but $f$ not Lebesgue Measurable.

Question

I have come up with a counterexample to the following statement:

Let $f:[0,1] \rightarrow \Bbb R$ and suppose that for every $c\in \Bbb R$, the set $\{x \in [0,1]\ |\ f(x)=c\}$ is $\mathcal L^1$-measurable. Then $f$ is $\mathcal L^1$-measurable.

Notation.

• $\mathcal L^1$ is the unique Carathéodory-Hahn extension of the volume function on disjoint intervals, which I call the Lebesgue measure.
• A set $A\subseteq \Bbb R$ is $\mathcal L^1$-measurable iff $\forall B \subseteq \Bbb R: \mathcal L^1(B) = \mathcal L^1(B\cap A) + \mathcal L^1(B\setminus A)$.
• A function $f$ is $\mathcal L^1$-measurable iff $\forall U \subseteq \Bbb R$ open, $f^{-1}(U)$ is $\mathcal L^1$-measurable.

Counterexample.

Let $P$ the Vitali set (or any uncountable non-measurable subset of $[0,1]$ such that $[0,1] \setminus P$ is uncountable). Let $f: [0,1] \rightarrow \Bbb R$ such that $f|_{P}$ is a bijection between $P$ and $\left[0,{1 \over 2}\right)$, and $f|_{[0,1]\setminus P}$ is a bijection between $[0,1]\setminus P$ and $\left[{1 \over 2}, 1\right]$.

Since $f: [0,1] \rightarrow [0,1]$ is a bijection, every $c\in \Bbb R$ has as a preimage either $\varnothing$ or a singleton $\{f^{-1}(c)\}$. Since singletons are Borel, they are $\mathcal L^1$-measurable and thus $f$ satisfies the above condition.

However the open set $\left[0,{1 \over 2}\right)$ has a non-measurable preimage, which means that $f$ is not $\mathcal L^1$-measurable. $\square$

Question. I know that for two uncountable sets $A, B$ there must exist a bijection between the two by transitivity, since for both there exists a bijection to $2^\omega$. However I would like to know if my construction is well-defined, assuming AOC of course, and how one would go about proving the existence of such a function rigorously. Furthermore is my reasoning correct that every $f^{-1}(c)$ is $\mathcal L^1$-measurable?

Note. For full disclosure, this is a true or false exercise from a homework problem sheet. However, the solutions don't give any proof/counterexamples to these true or false problems, which is why I'm asking here.

Answer 1

"I know that for two uncountable sets $A,B$ there must exist a bijection between the two by transitivity, since for both there exists a bijection to $2^ω$ ." Not really: first, you certainly assume $A,B⊆\Bbb R$ , and second, your claim that every uncountable subset of $\Bbb R$ is in bijection with $\Bbb R$ is equivalent to the continuum hypothesis. Better prove directly that there exist bijections between $P$ , $[0,1]\setminus P$ , and any non-trivial real interval. The rest is ok.

Of course, since $P$ is not given explicitely (its construction relies on AC), you won't be able to construct explicitly the required bijections. This doesn't prevent you from proving their existence.

For instance, $[0,1]=\sqcup_{p\in P}E_p$ where each $E_p$ is countable, hence $|[0,1]|=|P\times\Bbb N|$, which implies $|P|=|[0,1]|$. Similarly, $[0,1]\setminus P=\sqcup_{p\in P}(E_p\setminus\{p\})$ implies $|P|=|[0,1]\setminus P|$.

Answer 2

As noted by Simon Pitte, your function is a correct counterexample, but as Anne Bauval noted, proving that $P$ is in bijection with $[0,1/2)$ is tricky (even though not too difficult). I think this issue can be avoided, since you never needed a bijection anyway: an injection is enough. Define $f: [0,1] \to \mathbb{R}$ by $$f(x) = \begin{cases} \frac{1}{2}x, &x \in P,\\ \frac{1}{2}x + \frac{1}{2}, &x \in [0,1] \setminus P. \end{cases}$$ This $f$ is injective*, and hence every $f^{-1}(c)$ is still either empty or a singleton, so everything else is the same.

*: $f^{-1}(1/2)$ can possibly contain both $1 \in P$ and $0 \in [0,1] \setminus P$, but that's still finite and measurable. If you want, you can make $f|_{P}(x) = \frac{1}{2}x-\frac{1}{2}$ to avoid that.

Answer 3

Your function is well-defined: set-theoretically, a function $f:A\rightarrow B$ is formally defined as a triplet $(A,B,G)$, where $A$ and $B$ are sets, and $G$ (the graph of the function) is a subset of $A\times B$ such that for all $a\in A$, there is a unique $b\in B$ such that $(a,b)\in G$.

Assuming AC, $P$ exists, a bijection between $P$ and $[0,1/2)$ exists, and a bijection between $[0,1]\setminus P$ and $[1/2,1]$ exists. If you call $G_0$ the graph of that first function and $G_1$ the graph of the second, then the function you construct is, formally, the triplet $([0,1],\mathbb R,G_0\cup G_1)$ (this is the gluing of the two functions, which exists because unions exist by the axiom of union.) It’s a short exercise in formalism to show that this satisfies the definition of a function, as given above.

It’s also a short exercise in formalism to show it is injective, and hence the preimage of any point is indeed $\varnothing$ or a singleton, which are measurable.

Therefore your reasoning is correct, and your construction is in fact a counterexample to the statement given.