The standard probability measure over countably many independent coin tosses (i.e., the probability that you get a prescribed prefix of length $v$ is $2^{-v}$) is usually obtained via results in measure theory (at least, that's what I have seen). Is there a streamlined presentation out there that uses the least possible amount of results from measure theory (ideally, none) to show that this is indeed a valid probability measure?
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$\begingroup$ What does it mean to prove a statement whose very formulation requires measure theory without using any measure theory? $\endgroup$– Michael GreineckerCommented Jan 5, 2013 at 23:53
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This version is due to Emile Borel ...
Sequence of heads & tails encoded as 0s and 1s, then sequence is taken to represent a number in $[0,1]$ in its binary expansion. The measure is Lebesgue measure.
So you still need to know that Lebesgue measure exists.
answered Dec 31, 2012 at 22:25
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$\begingroup$ More in detail: let $D\subset 2^{\mathbb {N} _ +}$ be the countable set of eventually zero sequences $\{x_k \}_ {k\ge 1}$, which correspond to dyadic rationals of $[0,1]$, i.e. numbers with double binary representation. Then $ 2^{\mathbb {N} _ +}\setminus D\ni x\mapsto \sum_k\ 2^{-k}x_k \in (0,1]$ is a bi-measurable bijection. $\endgroup$ Commented Jan 1, 2013 at 12:29
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1$\begingroup$ Thanks, but I was hoping for something more basic. Real numbers are even more difficult than infinite coin tosses, in my opinion. How would you explain this probability measure to a high-school student? $\endgroup$– ManuCommented Jan 2, 2013 at 14:01
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$\begingroup$ Since this bijection goes both ways, each of Lebesgue measure and coin-tossing measure can easily be used to construct the other. Hence we should really not expect it to be any "easier" to construct one than the other. $\endgroup$ Commented Mar 29, 2019 at 16:45