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Nonnegative scalar multiples and countable weighted sums of measures are measures

Statement

Let $(μ_{k})$ be measures on one measurable space and let $c_{k} \in [0, + \infty]$ . Each scalar multiple $c_{k} μ_{k}$ , defined by the zero, finite-positive, and positive-infinity branches of Nonnegative scalar multiples and countable weighted sums of measures, is a measure. Every finite or countable weighted sum $\sum_{k} c_{k} μ_{k}$ is also a measure.

Statement references

Nonnegative scalar multiples and countable weighted sums of measures

Facts & Assumptions

Given: Measures $(μ_{k})$ on $(X, A)$ and coefficients $c_{k} \in [0, + \infty]$ .

[L1]

Scalar multiplication has separate $c = 0$ , $0 < c < + \infty$ , and $c = + \infty$ branches, and weighted sums are pointwise nonnegative extended sums (Nonnegative scalar multiples and countable weighted sums of measures).

[L2]

For a nonnegative extended double sequence, the two iterated sums are equal (Tonelli's theorem for double series of nonnegative extended real numbers).

[L3]

A measure vanishes at the empty set and is countably additive on disjoint measurable sequences (Measures on sigma-algebras).

Proof

technique · direct

1.1givenL1L3

For $c = 0$ , the set function $c μ$ is the zero measure.

1.2givenL1L3algebra

For $0 < c < + \infty$ , $(c μ) (\emptyset) = 0$ ; for disjoint $(E_{j})$ , multiplying the finite partial-sum identities by $c$ and taking their supremum gives $c μ (⋃_{j} E_{j}) = \sum_{j} c μ (E_{j})$ , whether the common value is finite or $+ \infty$ .

1.3givenL1L3

For $c = + \infty$ , a disjoint union has $μ$ -measure zero exactly when every member has $μ$ -measure zero: this follows directly from countable additivity and nonnegativity. Hence the infinite branch takes value $0$ on the union exactly when every term value is $0$ , and otherwise both it and the series of term values are $+ \infty$ ; this branch is a measure without forming $0 \cdot (+ \infty)$ .

2.1step 1.1step 1.2step 1.3

Steps 1.1, 1.2 and 1.3 prove that every scalar multiple $c_{k} μ_{k}$ is a measure for all possible coefficients.

3.1givenL1step 2.1

Put $ν (E) = \sum_{k} (c_{k} μ_{k}) (E)$ . Then $ν (\emptyset) = 0$ .

3.2step 2.1L1L2L3

If $(E_{j})$ is disjoint, then step 2.1 and Tonelli give $ν (⋃_{j} E_{j}) = \sum_{k} \sum_{j} (c_{k} μ_{k}) (E_{j}) = \sum_{j} \sum_{k} (c_{k} μ_{k}) (E_{j}) = \sum_{j} ν (E_{j})$ .

4.1step 3.1step 3.2L3∎

Steps 3.1 and 3.2 prove that the countable weighted sum is a measure; the same proof for a finite index range, or zero coefficients thereafter, gives every finite weighted sum, including the empty zero measure.

Depends on

Dependency tree · 2 levels

Sources

T. Tao, An Introduction to Measure Theory, Example 1.4.24 and Exercise 1.4.22