measure of Cantor type set

Question

Prove that the Lebesgue measure of the set $\{x\in[0,1]: \text{decimal expansion of $x$ contains only finitely many 7s}\}$ is zero.

I have thought that that if i can show that measure of $\limsup A_k$ is 1, then it is proven, where

$$A_k=\bigcup_i^{9^{k-1}} \left[\frac{10i+7}{10^k},\frac{10i+7}{10^k}\right]$$

($A_k:={}$set of $x$ such that $k$th decimal is $7$)

But by Borel Cantelli I have found that $\limsup A_k$ has measure zero. I think, the set in the questions is the complement of $\limsup A_k$. Any help or improvement or disproof of my statements are welcome.

I believe your union should go up to $10^{k-1}-2$? – Dustan Levenstein Nov 15 '13 at 23:56 — Dustan Levenstein, Nov 15 '13 at 23:56

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

3

Let $A$ be the set of numbers in $[0, 1]$ such that the decimal expansion doesn't contain any $7$s. By this question, $m(A) = 0$.

Let $B$ be the set of numbers in $[0, 1]$ such that the decimal expansion contains finitely many $7$s, and the rest are all $0$s. Since $B$ is a subset of $\mathbb Q$, it's countable.

Let $X$ be the set in your question. We have $$ X \subset \bigcup_{b \in B} (A + b). $$

By the translation invariance of the Lebesgue measure, we have $$ m(X) \le \sum_{b \in B} m(A + b) = 0. $$

edited Apr 13 '17 at 12:20

Community

1

answered Nov 16 '13 at 00:14

Ayman Hourieh

39,603

very cool :) thank you bro – seriously divergent Nov 16 '13 at 00:34
@Tony Happy to help! – Ayman Hourieh Nov 16 '13 at 00:35

measure of Cantor type set

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