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The simple integral is monotone, homogeneous, and additive

Statement

Let $s, t$ be nonnegative simple measurable functions and let $c \geq 0$ .

If $s \leq t$ pointwise, then $\int s d μ \leq \int t d μ$ .
If $c > 0$ , then $\int c s d μ = c \int s d μ$ . If $c = 0$ , then $\int 0 s d μ = 0$ . The second clause avoids forming the globally undefined extended-real product $0 \cdot (+ \infty)$ .
$\int (s + t) d μ = \int s d μ + \int t d μ$ .

Facts & Assumptions

Given: Nonnegative simple measurable functions $s, t$ and a scalar $c \geq 0$ .

[L1]

The simple integral is well defined, so any convenient common refinement of the chosen simple representations may be used to compute it (The simple integral is independent of the chosen representation).

[L2]

The simple integral of $\sum_{j} a_{j} χ_{E_{j}}$ is $\sum_{j} a_{j} μ (E_{j})$ with $0 \cdot (+ \infty) = 0$ (The integral of a nonnegative simple function).

Proof

technique · direct

1.1L1construct

Complete the representations of $s$ and $t$ with their zero-valued complements. Take their finite measurable common refinement $(E_{r})$ . On each cell write $s = a_{r}$ and $t = b_{r}$ . If $s \leq t$ , then $a_{r} \leq b_{r}$ .

1.2L2

The zero-scalar case is separate. [L2] When $c = 0$ , the function $c s$ is zero. Representing it by $0 χ_{X}$ gives $\int 0 s d μ = 0$ , even if $μ (X) = + \infty$ , by the definition's local zero-times-infinity convention.

2.1step 1.1L2

Monotonicity follows cell by cell. [step 1.1, L2] On the common partition, $\int s d μ = \sum_{r} a_{r} μ (E_{r}), \int t d μ = \sum_{r} b_{r} μ (E_{r}) .$ For finite or infinite $μ (E_{r})$ , the local simple-integral convention makes $a_{r} μ (E_{r}) \leq b_{r} μ (E_{r})$ whenever $0 \leq a_{r} \leq b_{r}$ . Summing these nonnegative extended-real inequalities proves clause 1.

2.2step 1.1L2

Additivity follows on the same partition. [step 1.1, L2] The coefficient of $s + t$ on $E_{r}$ is $a_{r} + b_{r}$ , and $(a_{r} + b_{r}) μ (E_{r}) = a_{r} μ (E_{r}) + b_{r} μ (E_{r})$ under the local zero-times-infinity convention. Finite sums in $[0, + \infty]$ can be regrouped without subtraction, so clause 3 follows.

3.1step 1.1L2∎

For $c > 0$ , scalar multiplication holds cell by cell. [step 1.1, L2] The identity $(c a_{r}) μ (E_{r}) = c (a_{r} μ (E_{r}))$ is valid in $[0, + \infty]$ for positive $c$ , and finite summation gives $\int c s d μ = c \int s d μ$ . Together with the preceding cases, this proves all three clauses.

Depends on

Used by

Additivity of the nonnegative Lebesgue integral Corollary
Monotonicity and nonnegative homogeneity of the nonnegative integral Proposition
The nonnegative integral agrees with the simple integral on simple functions Proposition
Monotone convergence for the integral Theorem
The indefinite integral of a nonnegative simple function is a measure Theorem

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John K. Hunter, Measure Theory Notes, Proposition 4.3 (standard reference, not scraped)