is sometimes used to illustrate the fact that the rational number
s are countable
while reals are not
. It is obviously true
. Considering that obviously true statements in probability theory
have a nasty habit of being false there is a point in proving it.
If X is a continuous random variable then P(X ∈ Q) = 0
Consider P(X = a). Take e > 0. Since X is a continuous random variable the function F(x) = P(X < x) is continuous. Therefore there exist d > 0 such that F(a+d) - F(a-d) < e. Thus P(X = a) < P(|X - a| < d) < e. Since e is an arbitrary positive number this implies P(X = a) = 0.
Let qn, n ∈ N be a listing of all rational numbers (there are such listings since Q is countable. Then
P(X ∈ Q) = P(UNION(n ∈ N)(X = qn)) = SUM(n ∈ N)(P(X = qn)) = 0
It is worth noting that the proof relies on the fact that the rationals are countable since a sum over an uncountable set would be meaningless.
A more direct approach would be to write P(X ∈ Q) = ∫ I[x ∈ Q]p(x)dx), where the integral is taken from -∞ to ∞. Unfortunately this is the classical example of a non-integrable function, so unless we opt for some non-Riemann definition of the integral this approach fails.