Let A be a countable subset of R. Show that m∗(A) = 0.
Since A is countable, we can write it as A = {a1, a2, a3, …}
For any δ > 0, we can cover each ai by an open interval Ii of length less than δ/2^i, as shown below:
[insert image]
Then, the set of intervals {Ii} covers A, and its total length is less than
∑|Ii| < δ
Therefore, m∗(A) ≤ δ. Since δ is arbitrary, we have m∗(A) = 0.
To show that the outer measure of a countable subset A of R is equal to zero, we'll need to use the definition of the outer measure and the properties of countable sets.
The outer measure of a set A, denoted by m∗(A), is defined as the infimum of the sums of the lengths of intervals that cover A. More formally, for any set A, we have:
m∗(A) = inf { Σ |I_n| : A ⊆ ⋃ I_n },
where I_n is a collection of intervals that cover A, and |I_n| represents the length of interval I_n.
Now, let's consider the countable set A ⊆ R. Since A is countable, it can be written as A = {a_1, a_2, a_3, ...}, where a_i represents the ith element of A.
To cover A, we can construct a collection of intervals I_n as follows: for each element a_i in A, we can choose an interval I_n_i = (a_i - ε/2^i, a_i + ε/2^i), where ε is a positive real number.
It is clear that A is contained within the union of all these intervals:
A ⊆ ⋃ I_n_i.
The length of each interval I_n_i is given by |I_n_i| = ε/2^i.
Now, we can calculate the sum of the lengths of these intervals:
Σ |I_n_i| = Σ ε/2^i,
where the summation is taken over all i.
The sum Σ ε/2^i is a convergent geometric series, since 0 < ε/2 < 1. Therefore, the sum converges to:
Σ ε/2^i = ε.
Since ε can be chosen arbitrarily, we can make it arbitrarily small. Hence, the infimum (i.e., the outer measure) is given by:
m∗(A) = inf { Σ |I_n| : A ⊆ ⋃ I_n } = 0.
Thus, we have shown that the outer measure of a countable subset A of R is equal to zero.