Non-Negative Additive Function is Monotone

Theorem

Let $\SS$ be an algebra of sets.

Let $f: \SS \to \overline \R$ be an additive function, that is:

$\forall A, B \in \SS: A \cap B = \O \implies \map f {A \cup B} = \map f A + \map f B$

If $\forall A \in \SS: \map f A \ge 0$, then $f$ is monotone, that is:

$A \subseteq B \implies \map f A \le \map f B$

Proof

Let $A \subseteq B$.

Then:

\(\ds B\)

\(=\)

\(\ds \paren {B \setminus A} \cup \paren {A \cap B}\)

Set Difference Union Intersection

\(\ds \)

\(=\)

\(\ds \paren {B \setminus A} \cup A\)

Intersection with Subset is Subset

\(\ds \O\)

\(=\)

\(\ds \paren {B \setminus A} \cap A\)

Set Difference Intersection with Second Set is Empty Set

\(\ds 0\)

\(\le\)

\(\ds \map f {B \setminus A}\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds \map f A\)

\(\le\)

\(\ds \map f {B \setminus A} + \map f A\)

\(\ds \leadsto \ \ \)

\(\ds \map f A\)

\(\le\)

\(\ds \map f B\)

Definition of Additive Function (Measure Theory)

$\blacksquare$