This statement is sometimes used to illustrate the fact that the rational numbers 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.

Taking this idea further, a random real is unspecifiable (almost always). Any number that has a representation, i.e. can be specified, is in a countable set - because the set of all mathematical representations is itself countable. (This idea can be taken further to say that Science is countable.) So a truly random number is unrepresentable.

More accurate would be to say that "a random real is irrational almost always".

After all, krimson's argument will show that for any real number y, P(X=y) = 0. Yet obviously X does take on some real value y, so we cannot say "X is never equal to y" -- we must say "X is equal to y with probability 0."

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