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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.

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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 $x \in K$.

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