Linear Combination of Measures

Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $\mu, \nu$ be measures on $\left({X, \Sigma}\right)$.

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


 * $a \mu + b \nu: \Sigma \to \overline{\R}, \ \left({a \mu + b \nu}\right) \left({E}\right) := a \mu \left({E}\right) + b \nu \left({E}\right)$

is also a measure on $\left({X, \Sigma}\right)$.

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:


 * $\forall E \in \Sigma: \left({a \mu + b \nu}\right) \left({E}\right) \ge 0$

Let $E \in \Sigma$.

Then $\mu \left({E}\right), \nu \left({E}\right) \ge 0$ as $\mu$ and $\nu$ are measures.

Hence, $a \mu \left({E}\right) \ge 0$ as $a \ge 0$.

Also, $b \nu \left({E}\right) \ge 0$ since $b \ge 0$.

Therefore it follows that:


 * $a \mu \left({E}\right) + b \nu \left({E}\right) \ge 0$

as desired.

Axiom $(2)$
Let $\left({E_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.

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


 * $\displaystyle \left({a \mu + b \nu}\right) \left({\bigcup_{n \mathop \in \N} E_n}\right) = \sum_{n \mathop \in \N} \left({a \mu + b \nu}\right) \left({E_n}\right)$

So let us do a direct computation:

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

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


 * $\left({a \mu + b \nu}\right) \left({\varnothing}\right) = 0$

This is verified by the following:

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

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