# What is the Lebesgue measure of the set of numbers in $[0,1]$ that has two thirds of ones in their infinite base-2 expansion?

Let $S= \{x \in [0,1 ]: \lim_{n \to \infty } \frac {1 } {n } \sum _{i=1 } ^n d _i (x) = 2/3 \}$, where $d _i(x)$ gives the ith digit of the infinite base-2 expansion of $x$. What is the Lebesgue measure of $S$?

First I want to prove that $S$ is a measurable set, and secondly determine its measure.

$d _i(X) : [0,1 ]\mapsto \{0,1 \}$,

$d _1^{-1 } (0)= [0,1)$
$d _1^{-1 } (1)=\{1\}$
$d _2^{-1 } (0) = [0,\frac {1 } {2 } )$...

So that $d _i$ is a measurable map? Now the sum is nondecreasing and bounded, hence converges pointwise. Thus the sum is a measurable map? Hence $S$ is measurable?

If my reasoning is correct, then what is the measure?

• Do you know that the $d_i$ are stochastically independent random variables? Do you know the (strong) law of large numbers? – PhoemueX Dec 13 '14 at 9:03
• Saw now that stochastically independent is just an other name for idependent... How can $d _i$ be seen as stochastic variables? – Alexander Dec 13 '14 at 10:39
• You interpret $([0,1], \mathcal{B}, \lambda)$ as a probability space, where $\mathcal{B}$ is the $\sigma$-algebra of Borel sets (or Lebesgue measurable sets if you wish) and $\lambda$ is Lebesgue measure. Then $d_i : [0,1] \to \{0,1\}\subset\Bbb{R}$ is a measurable map, i.e. a random variable. – PhoemueX Dec 13 '14 at 11:32