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?


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 $A$.


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