Let $A,F\subset\mathbb R$ be two sets such that $m(A)>0,\vert F\vert<\infty$. Prove $\exists\alpha,\beta\neq 0\quad \alpha+\beta F\subset A$

Question

Let $A,F\subset\mathbb R$ bw two sets, where $m(A)>0,\vert F\vert<\infty$
meaning $A$ has a positive Lebesgue measure, and $F$ has only finite number of points in it.

Prove $\exists\alpha,\beta\neq 0\quad \alpha+\beta F\subset A$

Where $\alpha+\beta F:=\{\alpha +\beta x \mid x\in F\}$

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My Attempt

We can assume WLOG that $F\subset [0,\varepsilon]$ for any $\varepsilon > 0$, and that $A\subseteq[-1,1]$.
(by applying the inverse of the isometry $x\mapsto \alpha + \beta x$ to $A$)

Using this:

$m(A)>0\implies\forall t\in(0,1)\quad \exists I\quad s.t\quad m(A\cup I)\geq t\cdot m(I) $

Or Lebesgue density theorem, I tried to show for some interval $J$ containing a density point of $A$ that $$\int\limits_J\chi_{A^c}(\alpha + \beta F)d\alpha = 0$$

Which proves $\chi_{A^c}\equiv0$ a.e, and therefore there is an $\alpha$ for which$\chi_{A^c}(\alpha + \beta F) = 0\implies \alpha + \beta F\subset J\cap A$.

I think this argument is wrong, because it can be used to prove the same statement with $F$ that is countable (edit- and bounded).

Is this argument wrong?
Is the statement is also true if $F$ is countable and bounded?

Thanks in advance for any help

Answer 1

Not true in the case: $A$ positive measure, and void interior ( a subset of Cantor type), and $F$ with non-void interior (like an open interval).

It's true if $F$ finite, $F=\{f_1, \ldots, f_k\}$.

First, for every $\epsilon >0$ there exists a segment $I$ so that $m(A\cap I) >(1-\frac{\epsilon}{2}) \cdot m(I)$. Now, take $\beta$ small enough so that $$m((A- \beta f_i) \cap I) >(1-\epsilon) \cdot m(I)$$ for all $1\le i \le k$ (if we move $A\cap I$ out of $I$ by a little bit, it won't change much the size of intersection with $I$)

Now, if $\epsilon$ is small enough ( say $\epsilon < \frac{1}{k}$ ) then the subsets $(A- \beta f_i) \cap I$ will have a non-void intersection (because the size of the union of their complements is $< m(I)$, and so does not cover $I$). Therefore, there exist $a_i \in A$ so that $a_i-\beta f_i$ are all equal for all $1\le i \le k$ $$a_i - \beta f_i = \alpha$$