Let fk:{0,1}∞→{0,1}kf_k :\{0,1\}^{\infty} \to \{0,1\}^k denote the projection map onto first kk component. Now given a probability measure PP on the measure space (Ω={0,1}∞,P(Ω))( \Omega=\{0,1\}^{\infty},P(\Omega)) we can push forward this probability measure using fkf_k and we obtain a probability measure on ({0,1}k,P({0,1}k)(\{0,1\}^k,P(\{0,1\}^k).

Suppose P1P_1 and P2P_2 are two probability measures on (Ω={0,1}∞,P(Ω))( \Omega=\{0,1\}^{\infty},P(\Omega)) such that the induced probability measure on ({0,1}k,P({0,1}k)(\{0,1\}^k,P(\{0,1\}^k) by P1P_1 and P2P_2 are same for each kk. Does this say that P1P_1 and P2P_2 are same measures on ({0,1}∞,P({0,1}∞))(\{0,1\}^{\infty},P(\{0,1\}^{\infty})) ?

I think that P1P_1 and P2P_2 should be some but i don’t have any formal argument.Any ideas?

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This is the uniqueness part of the Kolmogorov extension theorem.

– Nate Eldredge

Oct 20 at 22:34

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1 Answer

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The condition on P1\mathbb{P}_1 and P2\mathbb{P}_2 means that they agree on all cylinder sets, i.e. sets of the form

En={ω∈Ω:ω1=a1,…,ωn=an} E_n=\{\omega\in \Omega:\omega_1=a_1,\dots,\omega_n=a_n\}

where nn is a natural number and a1,…,an∈{0,1}a_1,\dots,a_n\in\{0,1\}. The cylinder sets (together with the empty set) form a π\pi-system which generates the product σ\sigma-algebra on Ω\Omega, and it follows from the π−λ\pi-\lambda theorem that if two probability measures agree on a π\pi-system, then they agree on the σ\sigma-algebra generated by this π\pi-system. Therefore P1=P2\mathbb{P}_1=\mathbb{P}_2.