# Question on measure of certain type of set

I have a question on the following problem:

Let $S\subseteq \mathbb{R}$ be a Borel set such that for any $s\in S$, if $s$ and $t$ differ in only a finite number of decimal places then $t\in S$. Then either $\lambda(S)=0$ or $\lambda(\mathbb{R}\setminus S)=0$.

I believe it has something to do with covering theorems, but my explorations in that regard have been unfruitful. I am also confused about the Borel hypothesis. I get the feeling therefore that I am overlooking some useful theorem.

As this is homework, I am looking for direction or advice and not a full answer. Thank you.

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OK. I think it is actually fixed now –  user51098 Nov 29 '12 at 4:14
Yes, that makes sense now. (I’ve not yet thought about proving it, however.) –  Brian M. Scott Nov 29 '12 at 4:17

My English is not perfect, so i am not sure what a decimal place is. Normally I would think it is any place behind the "dot" of a number. But in this case the open interval $]0,1[$ would be a counterexample to your statement. So I guess also things like $2 \in S \Rightarrow -3 \in S$ are valid. Assuming this:
The crucial idea is that the property of $S$ of being closed under finite changes of decimals implies a nice translation property: For all $t \in \mathbb R$ there is a sequence $t_n \to t$ such that $S + t_n \subseteq S$ for all $n$.
Assume $\lambda (S) \neq 0$. It is enough to show that $\lambda (S \cap [0,1]) = 1$. Since there is a point in $x \in S \cap [0,1]$ of density $1$ (see theorem 1 in the link above) it follows that $\lambda (B_r(x) \cap S)$ is arbitrary close to $2r$ for small $r$. Using the translation property things should work out.
@ Tony. Sorry, i don´t understand. (i changed $=$ to $\subseteq$, but i don´t think that´s what you mean) –  wspin Nov 29 '12 at 13:48
wspin, I edited the OP's question, not your answer. The original wording of the question suggested, perhaps, that $s$ and $t$ must differ in a finite, non-zero number of decimal places. –  TonyK Nov 29 '12 at 15:11