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This reasoning also shows the uniqueness of a nonnegative countably additive extension of µ to Aµ , since we have only used that A ∈ Aµ (however, as noted below, it is important that we deal with nonnegative extensions). A control question: where does the above proof employ the countable additivity of µ? 7. Example. Let A be the algebra of all ﬁnite subsets of IN and their complements and let µ equal 0 on ﬁnite sets and 1 on their complements. Then µ is additive and the single-point sets {n} cover IN, hence µ∗ (IN) = 0 < µ(IN).

5, taking into account that µ and µ coincide on A, we obtain µ∗ (A ∪ B) ≥ µ(Aε ∪ Bε ) − µ∗ (A ∪ B) (Aε ∪ Bε ) . 3) By the inclusion (A∪B) (Aε ∪Bε ) ⊂ (A Aε )∪(B Bε ) and the subadditivity of µ∗ one has the inequality µ∗ (A ∪ B) (Aε ∪ Bε ) ≤ µ∗ (A By the inclusion Aε ∩ Bε ⊂ (A Aε ) + µ∗ (B Aε ) ∪ (B µ(Aε ∩ Bε ) = µ∗ (Aε ∩ Bε ) ≤ µ∗ (A Bε ) ≤ ε. 4) Bε ) we have Aε ) + µ∗ (B ∗ Bε ) ≤ ε. ∗ Hence the estimates µ(Aε ) ≥ µ (A) − ε/2 and µ(Bε ) ≥ µ (B) − ε/2 yield µ(Aε ∪ Bε ) = µ(Aε ) + µ(Bε ) − µ(Aε ∩ Bε ) ≥ µ∗ (A) + µ∗ (B) − 2ε.

The next result called the monotone class theorem is frequently used in measure theory. 3. Theorem. (i) Let A be an algebra of sets. Then the σ-algebra generated by A coincides with the monotone class generated by A. 34 Chapter 1. Constructions and extensions of measures (ii) If the class E is closed under ﬁnite intersections, then the σ-additive class generated by E coincides with the σ-algebra generated by E. Proof. (i) Denote by M(A) the monotone class generated by A. Since σ(A) is a monotone class, one has M(A) ⊂ σ(A).