# 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 $\struct {X, \Sigma}$.

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: Add corollary for a positive linear combination of finitely many measuresYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## 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 $(2)$.

$\Box$

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### 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.

$\blacksquare$

## Sources

- 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $\S 4$: Problem $6 \ \text{(i)}$