Question

Let A be a countable subset of R. Show that m∗(A) = 0.

Answer 1

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.