$\mu(A \cap I) \le a \mu(I)$ implies $\mu(A) = 0$?

Question

Let $\mu$ be lebesgue measure on $\mathbb{R}$, $0<a<1$. If $\mu(A \cap I) \le a \mu(I)$ holds for any interval $I$, can I say $\mu(A)=0$?

I tried to construct a counterexample by considering something similar to cantor set but failed. Can anyone give some comments? Thanks!

Answer 1

Hint. By the Lebesgue density theorem, for a.e. $x \in \Bbb{R}$ we have

$$ \mathbf{1}_A (x) = \lim_{h \to 0^+} \frac{\mu(A \cap [x,x+h])}{\mu([x,x+h])}. $$

What can you conclude about the indicator function $\mathbf{1}_A$?

Answer 2

Suppose first $m(A) \subset [-R,R]$ for some $R>0.$ If $A \subset I_1 \cup I_2 \cup \cdots,$ where the $I_n$ are pairwise disjoint open intervals, then

$$\tag 1 m(A) = m(A\cap (\cup I_n))= m(\cup (A\cap I_n)) = \sum m(A\cap I_n) \le \sum am(I_n) = a\sum m(I_n).$$

Now the infimum of $\sum m(I_n)$ over all such converings of $A$ equals $m(A).$ Thus $(1)$ implies $m(A) \le a m(A).$ Because $m(A) < \infty,$ we must have $m(A) = 0.$

If $A$ is unbounded, then we can write $A = \cup_n\ A\cap [-n,n].$ Apply the above for each fixed $n$ to see each $m(A\cap [-n,n]) =0.$ Hence $m(A) = 0.$