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Let $Y$ denote the set of points of $[0,1)$ that have a unique binary expansion. Then $Y$ has a countable complement so $m(Y)=1$, where $m$ is the Lebesgue measure.

I have to confess that I do not know much about binary expansion and so I do not understand this assumption.

1.) I thought that every point in $[0,1)$ has a unique binary expansion. Why not?

2.) Why is the complement of $Y$ countable andwhy is then $m(Y)=1$?

Maybe you can help me to understand this.

Thank you

Adriano
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mathfemi
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1 Answers1

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(1) For instance: $0.01111\cdots=0.1$ in binary.

(2) All nonunique expansions are of the above form. In particular, the reals in $[0,1)$ with nonunique binary expansions are those that admit finite binary expansions. There are countably many of those.

Countable sets have Lebesgue measure zero.

anon
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    I do understand (1). Thank you. (2): Why are all nonunique expansions of that form? Why are the real numbers in $[0,1)$ with nonunique binary expansion those that admit finit binary expansions? And why are there countably many of those? - - Sorry thats not clear to me. – mathfemi Aug 21 '14 at 09:11
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    Assume a number $x$ has two distinct expansions, first differing at the $n$th place. Then my multiplying with $2^n$ and subtracting an integer, you get two representations $0.a_1a_2\ldots$ and $1.b_1b_2\ldots$ of the same number. The latter is $\ge 1$, the former is $\le 1$, hence both must equal $1$, so they must be $0.1111\ldots$ and $1.000\ldots$. Thus one of the original representations for $x$ ends in all zeroes, i.e. is finite. But a finite 0-1-string (ending in $1$) can be viewed as a reversed representation of an integer, hence there are only countably many – Hagen von Eitzen Aug 21 '14 at 09:26