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This question has stumped me.

Assume that $S\subseteq \mathbb{R}$ is meagre and that $S$ has the property that if $x\in S$ and $y$ differs from $x$ in only a finite number of decimal places then $y\in S$. Is it true that $\lambda(S)=0$?

I was thinking that differing by finitely many decimal places is an equivalence relation and noting that if $\lambda(S)>0$ it must be an uncountable union of these equivalence classes, but I could not get this to help me much.

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1 Answer 1

It is false. Let $S_0$ be any meagre measurable set of positive Lebesgue measure. For example, $S_0$ can be chosen as a fat Cantor set. Let $$F=\{x\in\mathbb{R}: \mbox{the decimal expansion of } x \mbox{ has only finitely many nonzero places} \}.$$ Then it is easy to see that $F$ is a countable set. Let $$S=\cup_{x\in F}(x+S_0)=\{x+s:x\in F,s\in S_0\}.$$ By definition, $S$ is measurable, $\lambda(S)>0$, and if $x\in S$ and $y$ differs from x in only a finite number of decimal places, then $y\in S$. Moreover, since $S$ is a countable union meagre sets, it is meagre.

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