The i.i.d. coin-flip measure is the uniform measure on its sample space
ProvedspMeasure_eq_uniformLet spMeasure n be the i.i.d. fair-coin product measure on SpOmega n = Fin n \to \mathrm{Bool}$ (the finite product of $n$ copies of the uniform measure on $\{\mathrm{true},\mathrm{false}\}$). Then spMeasure ncoincides exactly with the uniform probability measure on the finite typeSpOmega n`:
That is, a finite product of uniform (Bernoulli(1/2)) measures on a two-point space is itself the uniform (counting) measure on the product space: every one of the outcomes has exactly the same probability . This is the bridge that lets a measure-theoretic concentration bound proved on spMeasure n (e.g. via sub-Gaussian moment generating function machinery) be converted into a plain cardinality/counting statement about a Finset of outcomes, and vice versa.
Formalization Note. Proved via MeasureTheory.Measure.ext_of_singleton: both sides are probability measures on a finite (hence countable) type, so it suffices to check they assign the same mass to every singleton, which follows from Measure.pi_singleton and PMF.uniformOfFintype_apply after computing .
import Definitions.Def_Komlos_RandomSignModel import Mathlib open Komlos MeasureTheory
theorem spMeasure_eq_uniform (n : ℕ) :
spMeasure n = (PMF.uniformOfFintype (SpOmega n)).toMeasure := by sorry