Pluennecke inequality for Lebesgue measure?

Question

Suppose $G$ is a group. $A \subset G$. Let’s define $\{A_n\}_{n = 1}^\infty$ by the following recurrence:

$$A_n = \begin{cases} A && \quad n = 1 \\ A_{n - 1}A && \quad n > 1 \end{cases}$$

Here $AB$ means $\{ab| a \in A, b \in B\}$.

In group theory there is a following theorem:

Pluennecke inequality

Suppose $G$ is an abelian group and $A \subset G$ is a finite non-empty subset. Then $\forall n > 2$ $$|A_n| \leq \frac{|A_2|^n}{|A|^{n-1}}$$

I wonder if there is an analogical statement in real analysis. To be exact, my question is:

Is the following conjecture true?

My conjecture:

Suppose $A$ is a measurable subset of $\mathbb{R}$ ($\mathbb{R}$ is considered to be a group under addition), such that $\mu(A) > 0$ and $\mu(A_2) < \infty$. Then $\forall n > 2$ $$\mu(A_n) \leq \left(\frac{\mu(A_2)}{\mu(A)}\right)^n \mu(A)$$

Here $\mu$ stands for Lebesgue measure.

The additional supposition about the measure of $A_2$ is caused by this.

If that conjecture is false I would like to know an explicit counterexample.

Answer 1

Note that if $X$ has $\mu(X) > 0$ and $Y$ is unbounded, it's not hard to show that $\mu(X + Y) = \infty$ if $X+Y$ is measurable, so the assumption that $A_2 = A+A$ has $\mu(A_2) < \infty$ tells us that $A$ must be bounded. Define $M = \sup \{|a| : a \in A\}$, let $r = 3nM$, and consider the space $\mathbb{R}/r\mathbb{Z}$ with the measure given by regarding the space as $[0, r)$ and using the Lebesgue measure. We can then translate $A_1, \dots, A_n$ to the corresponding subsets of $\mathbb{R}/r\mathbb{Z}$ -- note that the quotient $\mathbb{R} \to \mathbb{R}/r\mathbb{Z}$ is injective on these sets, and preserves their measure, so it suffices to prove that the Plünnecke inequality holds for subsets of $\mathbb{R}/r\mathbb{Z}$.

This paper proves that the Plünnecke inequality holds for "$K$-analytic" subsets of compact abelian groups under the Haar measure, which applies in this case since $\mathbb{R}/r\mathbb{Z}$ is compact abelian and our choice of measure was a scaling of the Haar measure. This might be the best you can hope for, as the paper notes that the sum of Lebesgue measurable sets need not be Lebesgue measurable, but the class of $K$-analytic sets is still pretty large -- in the case of $\mathbb{R}/r\mathbb{Z}$ it includes the Borel sets.