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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?

flag	asked Dec 31 at 21:52 Emanuele Viola 193●5

What does it mean to prove a statement whose very formulation requires measure theory without using any measure theory? – Michael Greinecker Jan 5 at 23:53

Gerald Edgar · Answer 1 · 2012-12-31 22:25:04Z UTC

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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 at 22:25

Gerald Edgar
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	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. – Pietro Majer Jan 1 at 12:29
	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? – Emanuele Viola Jan 2 at 14:01

Streamlined probability measure for tossing infinitely many coins

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