Are these two measures same? Let $f_k :\{0,1\}^{\infty} \to \{0,1\}^k$ denote the projection map onto first $k$ component. Now given a probability measure $P$ on the measure space $( \Omega=\{0,1\}^{\infty},P(\Omega))$ we can push forward this probability measure using $f_k$ and we obtain a probability measure on $(\{0,1\}^k,P(\{0,1\}^k)$.

Suppose $P_1$ and $P_2$ are two probability measures on $( \Omega=\{0,1\}^{\infty},P(\Omega))$ such that the induced probability measure on $(\{0,1\}^k,P(\{0,1\}^k)$ by $P_1$ and $P_2$ are same for each $k$. Does this say that $P_1$ and $P_2$ are same measures on $(\{0,1\}^{\infty},P(\{0,1\}^{\infty}))$ ?

I think that $P_1$ and $P_2$ should be some but i don't have any formal argument.Any ideas?

The condition on $\mathbb{P}_1$ and $\mathbb{P}_2$ means that they agree on all cylinder sets, i.e. sets of the form 
$$ E_n=\{\omega\in \Omega:\omega_1=a_1,\dots,\omega_n=a_n\} $$
where $n$ is a natural number and $a_1,\dots,a_n\in\{0,1\}$. The cylinder sets (together with the empty set) form a $\pi$-system which generates the product $\sigma$-algebra on $\Omega$, and it follows from the $\pi-\lambda$ theorem that if two probability measures agree on a $\pi$-system, then they agree on the $\sigma$-algebra generated by this $\pi$-system. Therefore $\mathbb{P}_1=\mathbb{P}_2$.
