Question 10.T.13: If μ is any measure on B such that μ(I) = l(I) for all I ∈ I...
If μ is any measure on \mathcal{B} such that μ(I) = l(I) for all I ∈ I, then μ = m on \mathcal{B}.
Let E be any Borel set and suppose ε > 0. By Definition 10.8, there is a collection \left\{I_{i} ∈ \mathcal{I} : i ∈ \mathbb{N}\right\} such that E ⊆ \cup_{i=1}^{∞} I_{i} and
m(E) + ε ≥ \sum\limits_{i=1}^{∞}{l(I_{i})}= \sum\limits_{i=1}^{∞}{μ(I_{i})}.
Since μ is monotonic and countably subadditive,
\sum\limits_{i=1}^{∞}{μ(I_{i})} ≥ μ(E),
hence, ε being arbitrary,
m(E) ≥ μ(E). (10.13)
Suppose now that m(E) < ∞. Given ε > 0, choose \left\{I_{i} : i ∈ \mathbb{N}\right\} as above and assume, without loss of generality, that the union A = \cup_{i=1}^{∞} I_{i} is disjoint. Since m and μ are both countably additive, m(A) = μ(A); and since E ⊆ A,
m(E) ≤ m(A) = μ(A) = μ(E) + μ(A\E).
But μ(A\E) ≤ m(A\E) by (10.13), and m(A) ≤ m(E)+ε by assumption, hence
m(E) ≤ μ(E) + m(A\E)
= μ(E) + m(A) − m(E)
≤ μ(E) + ε,
which implies m(E) ≤ μ(E). In view of (10.13), we conclude that μ(E) = m(E).
We have therefore proved that μ(E) = m(E) for every E ∈ \mathcal{B} with m(E) < ∞. For a general E in \mathcal{B}, we can write
E = \underset{n∈\mathbb{Z}}{\cup} E ∩ [n, n +1) = \underset{n∈\mathbb{Z}}{\cup} E_{n},
where E_{n} = E ∩ [n, n + 1). Since, for each n, m(E_{n}) < ∞, we have μ(E_{n}) = m(E_{n}). The sets E_{n} being pairwise disjoint, the countable additivity of μ and m implies
m(E) = \sum\limits_{n∈\mathbb{Z}}{m(E_{n})} = \sum\limits_{n∈\mathbb{Z}}{μ(E_{n})} = μ(E).