# Linear Combination of Measures

## Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu, \nu$ be measures on $\struct {X, \Sigma}$.

Then for all positive real numbers $a, b \in \R_{\ge 0}$, the pointwise sum:

$a \mu + b \nu: \Sigma \to \overline \R, \ \map {\paren {a \mu + b \nu} } E := a \map \mu E + b \map \nu E$

is also a measure on $\struct {X, \Sigma}$.

## Proof

Verifying the axioms $(1)$, $(2)$ and $(3')$ for a measure in turn:

### Axiom $(1)$

The statement of axiom $(1)$ for $a \mu + b \nu$ is:

$\forall E \in \Sigma: \map {\paren {a \mu + b \nu} } E \ge 0$

Let $E \in \Sigma$.

Then $\map \mu E, \map \nu E \ge 0$ as $\mu$ and $\nu$ are measures.

Hence, $a \map \mu E \ge 0$ as $a \ge 0$.

Also, $b \map \nu E \ge 0$ since $b \ge 0$.

Therefore it follows that:

$a \map \mu E + b \map \nu E \ge 0$

as desired.

$\Box$

### Axiom $(2)$

Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.

The statement of axiom $(2)$ for $a \mu + b \nu$ is:

$\ds \map {\paren {a \mu + b \nu} } {\bigcup_{n \mathop \in \N} E_n} = \sum_{n \mathop \in \N} \map {\paren {a \mu + b \nu} } {E_n}$

So let us do a direct computation:

 $\ds \map {\paren {a \mu + b \nu} } {\bigcup_{n \mathop \in \N} E_n}$ $=$ $\ds a \map \mu {\bigcup_{n \mathop \in \N} E_n} + b \map \nu {\bigcup_{n \mathop \in \N} E_n}$ Definition of Pointwise Addition $\ds$ $=$ $\ds a \sum_{n \mathop \in \N} \map \mu {E_n} + b \sum_{n \mathop \in \N} \map \nu {E_n}$ $\mu$ and $\nu$ are measures and satisfy $(2)$ $\ds$ $=$ $\ds \sum_{n \mathop \in \N} a \map \mu {E_n} + b \map \nu {E_n}$ Combined Sum Rule for Real Sequences $\ds$ $=$ $\ds \sum_{n \mathop \in \N} \map {\paren {a \mu + b \nu} } {E_n}$

which establishes $a \mu + b \nu$ satisfies $(2)$.

$\Box$

### Axiom $(3')$

The statement of axiom $(3')$ for $a \mu + b \nu$ is:

$\map {\paren {a \mu + b \nu} } \O = 0$

This is verified by the following:

 $\ds \map {\paren {a \mu + b \nu} } \O$ $=$ $\ds a \map \mu \O + b \map \nu \O$ Definition of Pointwise Addition $\ds$ $=$ $\ds a \cdot 0 + b \cdot 0$ $\mu$ and $\nu$ are measures and satisfy $(3')$ $\ds$ $=$ $\ds 0$

Thus, $a \mu + b \nu$ satisfies $(3')$.

$\Box$

Having verified an appropriate set of axioms, it follows that $a \mu + b \nu$ is a measure.

$\blacksquare$