Let {rn} be an enumeration of Q ∩ [0,1] and let > 0. Let B = [0,1] \ A, where A= n=1 (b) Let A =1 2n,rn + 2n. Find the Lebesgue outer measure of B.
Since {rn} is an enumeration of Q ∩ [0,1], we have B = [0,1] \ Q ∩ [0,1].
For any given ε > 0, we can cover Q ∩ [0,1] by a countable collection of open intervals {In} with total length < ε/2. Then B is contained in the union of the complementary intervals {Jn = (a_n, b_n)} where a_n and b_n are the endpoints of In. Each Jn is a closed interval and has length l(Jn) = b_n - a_n > 0.
Notice that the sets {Jn} and {I_n} form a countable collection of open intervals that cover [0,1]. By the countable additivity of Lebesgue measure, we have
1 = m([0,1]) ≤ ∑n m(I_n) ≤ ∑n (l(Jn) + ε/2) = ∑n l(Jn) + ε/2.
Thus, we have ∑n l(Jn) ≥ 1 - ε/2. Since B ⊆ ∪n Jn, we have
m*(B) ≤ ∑n l(Jn) ≤ 1 - ε/2.
Since ε is arbitrary, we conclude that m*(B) ≤ 1.
To show that m*(B) ≥ 1, it suffices to show that for any δ > 0, there is an open interval U = (a,b) containing [0,1] \ B with m(U) ≤ 1 + δ. To see this, notice that we can then cover U ∩ Q ∩ [0,1] by a countable collection of open intervals {In} with total length < δ/2. The set B is contained in the complement of this cover, which is the union of the intervals {Jn = (a_n, b_n)} where a_n and b_n are the endpoints of In. Each Jn is a closed interval and has length l(Jn) = b_n - a_n > 0. Then,
m*(B) ≤ ∑n_l(Jn) ≤ m(U) + δ/2 ≤ 1 + δ.
Thus, m*(B) ≥ 1 by the arbitrariness of δ.
Combining the upper and lower bounds, we have m*(B) = 1.
To find the Lebesgue outer measure of B, we need to determine the measure of the set B.
Given that B = [0,1] \ A, we need to calculate the measure of A and then subtract it from the measure of [0,1].
To find the measure of A, observe that A is the union of intervals (rn, rn + 2^n) for n = 1, 2, 3, .... Since each of these intervals has length 2^n, we can calculate the measure of A as the sum of the lengths of these intervals.
Measure(A) = Σ_n=1 to ∞ (2^n) = 2 + 4 + 8 + ... = 2(1 + 2 + 4 + ...) = 2 * (2/1) = 4.
The measure of [0,1] is simply 1, as it is the length of the interval [0,1].
Therefore, the measure of B is given by Measure([0,1]) - Measure(A) = 1 - 4 = -3.
However, the Lebesgue measure is non-negative and cannot be negative, so the Lebesgue outer measure of B is 0.
Hence, the Lebesgue outer measure of B is 0.