- Show that a countable union of null sets is null.
Solution
Let \((N_n)_{n\ge1}\) be measurable sets with \(\mu(N_n) = 0\) for all \(n\).
1. \(\mu\!\left(\bigcup_n N_n\right) \leq \sum_n \mu(N_n)\).
This is countable subadditivity of the measure \(\mu\).
2. Q.E.D.
Each term of \(\sum_n \mu(N_n)\) is \(0\), so step 1 gives \(\mu(\bigcup_n N_n) \leq 0\). Since \(\mu\) is nonnegative, \(\mu(\bigcup_n N_n) = 0\).