Math Forum / Mathematics / Undergraduate Math / February 2005

You might like to read up on measure spaces - these generalise the notion of
area to a wider range of "measurable sets".  In this context, a probability
space is a measure space with total measure 1.  Probability "events" are
measurable sets in the probability space, with the probability of the event
being the measure of the set.

For your interval [0,1], a probability (measure) space can be constructed
along these lines:
a) for a subset like [a,b) of [0,1], it's measure is defined as (b-a)
b) this implies the measure of a whole load of new sets, by taking countable
unions and differences of other sets
c) then using a concept of "outer measures", a satisfactory measure can be
defined on a wide class of subsets of [0,1] which is consistent with
previous steps (a) and (b).
(This whole process is quite fiddly, I recall, so I'm just giving a rough
impression of the steps here - don't ask me to explain what an outer measure
is, as I haven't looked at this for over 20 years... ;-)
d) the measure defined in (c) is is the "Lebesgue" measure space on [0,1],
which can be taken as your uniformly random distribution on [0,1].
Interestingly, not all subsets of [0,1] are measurable and so for certain
sets it would make no sense to ask for the probability that a randomly
selected x lies in that set...  However all intervals (and unions thereof)
and singleton sets are measurable - in fact you would have to think quite a
bit to construct an unmeasurable set...

In this probability space, P( {x} ) = 0 for every number x.  And so for any
finite or countable set C we also have P(C) = 0 (basically because measure
spaces obey rules e.g. that measures of disjoint sets add up to give the
measure of the union).

Yes, for any finite subset S of [0,1] will have P(S) = 0.
So P(x,n) = 0 for each n, and the limit is obviously zero too.  Also as
noted above, a *countable* subset C of [0,1] will still have P(C) = 0, which
is not saying the same thing as your result.

Unfortunately there's a problem here - if we could do the same as with the
Reals in [0,1] then the same argument would lead to P( {q} ) = 0, for each
rational q in [0,1].  But there are only countably many such q, and so the
total measure of the measure space would be zero, not 1.  I.e. constructing
a uniform distribution on just the rationals cannot work.

Well, this doesn't work the same way, because we cannot construct a uniform
measure in the way you want.  However, there are many ways of defining a
probability space on top of the set of rationals in [0,1].  If we enumerate
the set as q1, q2, ..., we can assign discreet probabilities p1, p2, ... to
each rational, as long as p1 + p2 + ... = 1.  (I.e. the sum of the infinite
series is 1, so the probability of the whole set is 1 as required for any
probability space).

Note such a probability space couldn't be classified as "uniformly
distributed"...

Now with your notation above, P(x,n) could be anywhere between 0 and 1
inclusive, depending on the distribution and the particular n rationals
we've chosen.  Similarly the limit could be anything between 0 and 1
inclusive.

One thing we can say for sure, if we have any sequence of rationals q1, q2,
q3,... such that
   {rationals in [0,1]} = {q1, q2, q3, ...} and we define
   P_n = P( {q1, q2, ... qn} )

then we will have Lim{n->oo: P_n} = 1

I can't see an interpretation for it in this context.  When discussing
ordinals in set theory there is a concept of "limit ordinals" beyond the
first infinite ordinal, but that's another story... :)

Hope the above is of interest to you,
Mike.