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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?

Thanks in advance!

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  • $\begingroup$ Do you know that the $d_i$ are stochastically independent random variables? Do you know the (strong) law of large numbers? $\endgroup$ – PhoemueX Dec 13 '14 at 9:03
  • $\begingroup$ I know the strong law of large numbers, not what stochastically independent means, but please refere me and I will read... $\endgroup$ – Alexander Dec 13 '14 at 9:11
  • $\begingroup$ Saw now that stochastically independent is just an other name for idependent... How can $d _i $ be seen as stochastic variables? $\endgroup$ – Alexander Dec 13 '14 at 10:39
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    $\begingroup$ 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. $\endgroup$ – PhoemueX Dec 13 '14 at 11:32
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    $\begingroup$ See this question math.stackexchange.com/questions/1067585/…. $\endgroup$ – PhoemueX Dec 14 '14 at 13:02

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