Now, let Pu be the uniform distribution on B∞, which samples infinite binary sequences one bit at a time, each with probability 50% to be 0 or 1.
Pu as defined here can’t be a proper/classical probability distribution over B∞ because it assigns zero probability to every s∈B∞: Pu(s)=limn→∞∏ni=1P(si=1)=limn→∞∏ni=112=limn→∞12n=0.
It’s a continuous probability measure, meaning it has no atoms, but it does assign positive probability to all cylinder sets. If you take the binary representation of reals in the [0,1] interval, P_u comes from the Lebesgue measure (a uniform distribution).
Pu as defined here can’t be a proper/classical probability distribution over B∞ because it assigns zero probability to every s∈B∞: Pu(s)=limn→∞∏ni=1P(si=1)=limn→∞∏ni=112=limn→∞12n=0.
Or am I missing something?
It’s a continuous probability measure, meaning it has no atoms, but it does assign positive probability to all cylinder sets. If you take the binary representation of reals in the [0,1] interval, P_u comes from the Lebesgue measure (a uniform distribution).
So the probability of a cylinder set Γ(x)={s∈B∞∣x⪯s} is 2−l(x) etc?
Yes.