By Moore E.H.

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An ∈ A are disjoint, then µ( µ is monotone and finitely subadditive. i Ai ) = i µ(Ai ). It follows that 3. Also, if A1 , A2 , . . ∈ A are disjoint, and i Ai happens to be in A, the previous equation must also hold. ) Then we have the following important result. 4. A ⊆ M, and µ∗ (A) = µ(A) for all A ∈ A. Thus µ∗ is a measure extending of µ onto the sigma algebra M containing the algebra A. ) Proof. Fix A ∈ A. For any E ⊆ X and ε > 0, by definition we can find A1 , A2 , . . ∈ A with E ⊆ n An and n µ(An ) ≤ µ∗ (E) + ε.

Also let d(x, A) = inf y∈A d(x, y) be the distance from x ∈ X to A. Set Dn = {x ∈ X : d(x, A) ≥ 1/n}. Dn is closed, because d(·, A) is a continuous function, and [1/n, ∞] is closed. Clearly d(x, A) ≥ 1/n > 0 implies x ∈ Ac = B, but since A is closed, the converse is also true: for every x ∈ Ac = B, d(x, A) > 0. Obviously the Dn are increasing, so we have just shown that they in fact increase to B. Hence µ(B \ Dn ) < ε for large enough n. Thus B ∈ M. 44 The case that µ is not a finite measure is taken care of, as you would expect, by taking limits like we did for sigma-finite measures in Section 8.

E) = i α−1 i (E∩Vi ) It is left as an exercise to show that ν(E) is well-defined: it is independent of the coordinate charts αi used for M . Finally, the scalar integral of f : M → R over M is simply f dν . M And the integral of a differential form ω on an oriented manifold M is ω p ; T (p) dν , p∈M where T (p) is an orthonormal frame of the tangent space of M at p, oriented according to the given orientation of M . ) Again it is not hard to show that the formulae I have given are exactly equivalent to the classical ones for evaluating scalar integrals and integrals of differential forms, which are of course needed for actual computations.