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?

$\begingroup$ What does it mean to prove a statement whose very formulation requires measure theory without using any measure theory? $\endgroup$ – Michael Greinecker Jan 5 '13 at 23:53
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.

$\begingroup$ 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 bimeasurable bijection. $\endgroup$ – Pietro Majer Jan 1 '13 at 12:29

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 highschool student? $\endgroup$ – Manu Jan 2 '13 at 14:01

$\begingroup$ Since this bijection goes both ways, each of Lebesgue measure and cointossing 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. $\endgroup$ – Nate Eldredge Mar 29 '19 at 16:45