 Solution

ID: past-exam-of-the-integration-course-of-the-university-of-oxford/2015/1/b/solution

For the increasing family, define the measurable sets

D_{1} = A_{1}, D_{n} = A_{n} ∖ A_{n - 1} (n \geq 2) .

They are pairwise disjoint, with

A_{N} = ⋃_{n = 1}^{N} D_{n}

and

⋃_{n} A_{n} = ⋃_{n} D_{n}

μ (⋃_{n = 1}^{\infty} A_{n}) = \sum_{n = 1}^{\infty} μ (D_{n}) = lim_{N \to \infty} \sum_{n = 1}^{N} μ (D_{n}) = lim_{N \to \infty} μ (A_{N}) .

The nonnegative series is defined by its increasing partial-sum limit, so this proof also covers an infinite value. This is continuity from below of a measure.

For the decreasing family, put

C_{n} = B_{1} ∖ B_{n}

. These measurable sets increase, and

⋃_{n} C_{n} = B_{1} ∖ ⋂_{n} B_{n} .

Since

μ (B_{1}) < \infty

μ (C_{n}) = μ (B_{1}) - μ (B_{n}), μ (B_{1} ∖ ⋂_{n} B_{n}) = μ (B_{1}) - μ (⋂_{n} B_{n}) .

Apply the first result to

C_{n}

and subtract from

μ (B_{1})

to obtain

μ (n = 1 ⋂ \infty B_{n}) = n \to \infty lim μ (B_{n}) .

B_{n} = [n, \infty)

, every

B_{n}

has infinite measure but their intersection is empty. The proof must not subtract

\infty - \infty

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