0
$\begingroup$

I was wondering about the following result: We are given a subset of the reals $A$ such that $A + r = A$ for some real number $r$ (in my application I have in mind for each $r$ belonging to a dense subset of the reals). Then assuming $A$ has a positive non-zero Lebesgue measure it follows that $A$ has infinite Lebesgue measure.

My approach: From the invariance we get a disjoint countable cover by sets $A_n := A \cap [(n-1) r, nr)$. But I don't see how to argue that for infinitely many $n$ the sets $A_n$ have positive Lebesgue measure which would clinch the deal. Ideas?

The background why I thought this is true (and should be easy to prove): In general there is a Theorem like this which is more difficult to prove without the assumption that $A$ be measurable in the first place. Namely if $A$ is translation invariant wrt a dense subset then $A$ is either of infinite measure or immeasurable (if I remember correctly).

EDIT: Additional question: What happens if instead of Lebesgue measure we consider a measure on the real numbers which is not necessarily translation invariant?

$\endgroup$
1
$\begingroup$

Assume $A \cap [a,b]$ has positive measure then chose $n$ such that $$((A\cap[a,b])+nr)\cap [a,b]=\emptyset$$ so the sets $(A\cap [a,b])+knr$ are all disjoint of equal positive meaasure and subsets of $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.