# If there is a zero measure set with positive measure pre-image, is there a level set with positive measure?

Let $$f:\mathbb{R}\to\mathbb{R}$$ be smooth (I suppose $$C^1$$ should be enough, but I would be happy with an answer when $$f\in C^\infty$$). Assume that there exists a Borel set $$A\subset\mathbb{R}$$ with zero Lebesgue measure whose pre-image $$f^{-1}(A):=\{x\in\mathbb{R}:f(x)\in A\}$$ has positive Lebesgue measure. Does this imply that there exists a point $$a\in\mathbb{R}$$ whose preimage $$f^{-1}(\{a\})$$ has positive measure?

Note that our condition implies that $$\{x\in\mathbb{R}:f'(x)=0\}$$ has positive measure (as otherwise one can rule out the existence of such a set $$A$$; see, for instance, this question). However there are smooth strictly increasing functions where $$\{x\in\mathbb{R}:f'(x)=0\}$$ is a Cantor set with positive measure.

There is a $$C^\infty$$ counterexample.

For $$x \in [0,1]$$, let $$\psi_0(x)=\int_0^x\exp\left(- \, \frac1{t(1-t)}\right)\,dt \,.$$ Then the strictly increasing function $$\psi \in C^\infty [0,1]$$ defined by $$\psi(x)=\psi_0(x)/\psi_0(1)$$ satisfies $$\psi(0)=0$$, $$\psi(1)=1$$, and all the derivatives of $$\psi$$ vanish at $$0$$ and at $$1$$.

Let $$K$$ be a fat cantor set, obtained by removing the middle $$1/4$$ from the unit interval, then the middle $$1/9$$ from each of the two resulting stage $$1$$ intervals, and in general, removing the middle $$1/(k+2)^2$$ from each of the $$2^k$$ intervals obtained at stage $$k$$. Clearly $$K$$ has positive measure, and each of the stage $$k$$ intervals in its construction has length $$2^{-k} \gamma_k$$, where $$\gamma_k:=\prod_{j = 0}^{k-1} \Bigl(1-\frac1{(j+2)^2}\Bigr) \to \gamma>0 \,.$$

Let $$\Lambda$$ be a thin cantor set, obtained by removing the middle $$1/3$$ from the unit interval, then the middle $$1/2$$ from each of the two resulting stage $$1$$ intervals, and in general, removing the middle $$(k+1)/(k+3)$$ from each of the $$2^k$$ intervals obtained at stage $$k$$. Each of the stage $$k$$ intervals in this construction has length $$\frac{2^{1-k}}{(k+2)!}$$.

We start by defining $$f:K \to \Lambda$$ that for every $$k$$, maps each of the $$2^k$$ stage $$k$$ intervals in the construction of $$K$$ to the corresponding interval in the construction of $$\Lambda$$. The complement $$[0,1] \setminus K$$ is a disjoint union of open intervals. Extend $$f$$ to each of these complementary intervals $$(a,a+r)$$ by $$\forall x \in (0,r), \quad f(a+x)=f(a)+\bigl(f(a+r)-f(a)\bigr)\cdot \psi(x/r)\,.$$ These extensions yield a strictly increasing function $$f \in C[0,1]$$ which satisfies $$f^{-1}(\Lambda)=K$$.

Our choice of $$\psi$$ implies that $$f$$ is $$C^\infty$$ on $$[0,1]\setminus K$$.

To show that $$f \in C^{\infty}[0,1]$$, one can verify by induction on $$\ell$$ that for all $$\ell \ge 1$$,

$$(*) \quad f \in C^\ell[0,1]$$ with $$f^{(\ell)}(x)=0$$ for all $$x \in K$$.