Proof to motivate the need for measure theoretic probability

Question

Im using Billingsley's Probability and Measure. In lecture the instructor motivated the need for measure theory in probability by providing a solution to the following problem:

Show that a discrete probability space cannot contain an infinite sequence $A_1, A_2,...$ of independent events, each of probability $\frac{1}{2}$. This is exercise 1.1a in the text.

My thought is that such a sequence may be interpreted as a dyadic expansion of some real number in the unit interval, thus the probability space must be unaccountably infinite. Of course this is not rigorous and it differs from the approach suggested by the author. The problem notes give the following hint:

"Each point lies in one of the four sets $A_1 \cap A_2$, $A_1^c \cap A_2$, $A_1 \cap A_2^c$, $A_1^c \cap A_2^c$ and hence would have probability at most $2^{-2}$; continue."

I'm not sure I see where he is going with this. Of course as the sequence goes infinite, the probability of any set goes to zero, but how does this prove the space cannot be discrete?

Answer 1

Let $x$ be an element of the probability space $X$. Let $p$ be the probability of the event $\{x\}$. Consider the $2^n$ pairwise disjoint sets of the form $A_1^{\varepsilon_1}\cap\dots\cap A_n^{\varepsilon_n}$ where for each $i\in\{1,\dots,n\}$, $\varepsilon_i\in\{0,1\}$, $A_i^0=A$ and $A^1_i=X\setminus A$. Each of these sets has probability $2^{-n}$ and $x$ is an element of one of them. It follows that $p\leq 2^{-n}$. Since this holds for every $n$, $p=0$. So every singleton in your space has probability $0$ and hence the space is not discrete.

Answer 2

Depends on what is meant by a discrete space but I presume its one for which there there has to be some event with positive probability, and there cannot be a continnuum of event. There are generally at most countably many atomic events with positive probability, if the space is uncountable; as normalization would fail.

In this case, it would fail regardless, every sequence has the same probability; they all must have probability zero if any one of them does. If countable addivity is not to violate normalization , all of the events must have PR=0 have probability zero (as any given sequence of prob 1/2 have the same probability, if one has Pr=0, they all do, they all do) . Lest independence or equiprobability Pr(A)=1/2) fails; if this fact alone does not constitute the impossibility of the space being discrete; as no positive valued events, then the fact that one could add them up nonetheless,as there are only countably many such events; then PR(sample space)=0, normalization would be violated. If they all had positive probability Pr(sample)=infinite (which they dont), and it still would be violated. If some of them had positive probability and other sequences did not (it may be possible) but then the sequences would not be equi-probable, or independent.

(1)One cannot have a countably infinite number of equi-probable events, if countable addivitity is to hold, without all of said events either having zero probability or positive probability in which Pr(sigma)=infinite, or Pr(sigma)=0; thus even if this is avoided by assigning such events zero probability;

(2)if discrete spaces allow for all events to have zero probability, Pr(sigma)=0. So it would not matter, even if a discrete space allowed for every sequence to have zero probability, normalization would still fail, as every sequence would have pr=0. And as there countably many of them, would use countable addivitity.

1.Either they must all have zero probability but there must be un-countably such events, so that one cannot derive a contradiction via normalization, via countable addivity,

2. Or they must have positive probability, and normalization will fail as there are only countably many events in the space. And they will add to more than one. Or they will not be independent, to begin with, as their probability must limit to zero, and if one sequence does, they all do (being equi-probable) and so Pr(sigma)=0, as every sequence has the same probability value.

Otherwise, the events will not be independent. Not that some approach such as frequentists try to avoid this issue, by postulating certain convergence. I am wondering if this can be avoiding by dropping countable addivitity; or using non-standard analysis.

(as there is no axiom of uncountable addivity); or if said events had positive probability (not only would this violate the laws of the probability calculus),

. And each sequence of independent and identical distributed events, is equally probable. One can in other cases,

By discrete; Do you mean that at there are at most countable infinitely, many events in F, the sigma algebra. Or that every event has positive probability. Every event is an atom. It is well known that every sequence with probability 1/2 has the same probability as each other, and if one has a probability of zero, every sequence will. I am wondering if this generalizes; it known that events when not equi-probable, can have positive probability, if there are only countable infinitely many of them (but in this case even if there are only countably infinitely many such sequences, they are all equally probable, and so that route of escape is not viable

Moreover,if these events are independent, then there will be un-countably many such sequences if there is one, in the sigma algebra. Otherwise certain sequences will be ruled and will not be independent; In particular probability half events converge much more slowly then other events (the probability of attaining any given sequence with probability close to the relative frequency) is lower then for other probability values.

The strong law of large numbers holds for such events, because there are simply much more of them. It is, also well known that almost all (uncountably many sequences, and most such sequences) by a theorem of Borel (normality theorem) consisting of binary trials, from amongst the uncountably infinite totality) have relative frequency 1/2.

Then this follows. Every infinite sequence of of probability 1/2 has probability of limiting to zero; Countable addivity would either fail, fail as P(sigma)=0 and likewise the law of large numbers which follow from it.

(2)To ensure that that one does get a violation of normalization there cannot be only countably infinitely many such events.

Lest one could add them up, vis a vis countable additivity; derive the failure of normalization.

To get around this issue; countable addivitity cannot apply directly and to ensure this, one needs an uncountable number of such sequences

Once cannot have a countable infinite many equiprobable events (infinite sequences of probability 1/2) much less an un-countably many, unless there probability is zero. (3) or otherwise the events could not be independent.

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And in this case, the case, the measure and the proportion must line up (by the strong law of large numbers); but almost all the measure is not on said sequence (they add to zero) in a discrete space; almost all such sequences do not limit to relative frequency half, and as each sequence is as likely as another; it is not the case that almost surely, relative frequency convergence to the probability value will occur. I once posed this question to my maths professorm in relation to whether there are uncountably many countably infinite sequences with an irrational value relative frequency value.

There are.

Otherwise it would be a counter-example to the strong law of large number, failing for such values as well, even, in a non discrete space. For the same reason. One could add their probability values, and attain that the probability that such a sequence would converge would be close to zero