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Prove. The measure of the set R of all rational numbers in the interval [0, 1] is zero.

The proof of the theorem utilizes the following theorem.

Theorem 1.

Proof. The derivation of 1) is as follows: The progression

is a geometric progression a, ar, ar^{2}, ar^{3} , .... where a = 1 and r = 1/2.. The sum of 2) over n
terms

is given by the formula

The sum of the progression 3) for n = is then given by

Thus, from 3),

Proof of main theorem: The measure of the set R of all rational numbers in the interval [0, 1] is zero.

We have already shown that the set R is countable. We now arrange the points of R in a sequence

r_{1}, r_{2}, ... , r_{n}, ... .

Next, for a given ε > 0, we enclose the point r_{n} by an open interval C_{n} of length ε/2^{n}. The sum C
= Σ C_{n} is an open covering of set R. The open intervals C_{n} may intersect so the measure mC of C
is given by

mC = m(Σ C_{n}) < Σ mC_{n} = Σ (ε /2^{n}) = ε

(where in the last equality we employ Theorem 1). Since ε can be chosen arbitrarily small, we have mR = 0, which is what we set out to prove.

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