Series of Measures is Measure

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Theorem

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

Let $\sequence {\mu_n}_{n \mathop \in \N}$ be a sequence of measures on $\left({X, \Sigma}\right)$.

Let $\sequence {a_n}_{n \mathop \in \N} \subseteq \R_{\ge 0}$ be a sequence of positive real numbers.


Then the series of measures $\mu: \Sigma \to \overline \R$, defined by:

$\ds \map \mu E := \sum_{n \mathop \in \N} a_n \map {\mu_n} E$

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


Proof

Let us verify the conditions for a measure in turn.


Let $E \in \Sigma$.

Then:

$\forall n \in \N: \map {\mu_n} E \ge 0$

and so:

$a_n \map {\mu_n} E \ge 0$

Therefore, by Series of Positive Real Numbers has Positive Limit:

$\ds \map \mu E = \sum_{n \mathop \in \N} a_n \map {\mu_n} E \ge 0$


For every $n \in \N$, also $\map {\mu_n} \O = 0$.

Therefore, it immediately follows that:

$\ds \map \mu \O = \sum_{n \mathop \in \N} a_n \map {\mu_n} \O = \sum_{n \mathop \in \N} 0 = 0$


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

Then:

\(\ds \map \mu {\bigcup_{m \mathop \in \N} E_m}\) \(=\) \(\ds \sum_{n \mathop \in \N} a_n \map {\mu_n} {\bigcup_{m \mathop \in \N} E_m}\) Definition of $\mu$
\(\ds \) \(=\) \(\ds \sum_{n \mathop \in \N} a_n \paren {\sum_{m \mathop \in \N} \map {\mu_n} {E_m} }\) The $\mu_n$ are measures
\(\ds \) \(=\) \(\ds \sum_{n \mathop \in \N} \sum_{m \mathop \in \N} a_n \map {\mu_n} {E_m}\)
\(\ds \) \(=\) \(\ds \sum_{m \mathop \in \N} \sum_{n \mathop \in \N} a_n \map {\mu_n} {E_m}\) Tonelli's Theorem: Corollary
\(\ds \) \(=\) \(\ds \sum_{m \mathop \in \N} \map \mu {E_m}\) Definition of $\mu$


Therefore, having verified all three axioms, $\mu$ is a measure.

$\blacksquare$


Sources

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