Linear Combination of Signed Measures is Signed Measure

Theorem

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

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

Let $\alpha, \beta \in \overline \R$.

Suppose that the sum:

$\alpha \map \mu A + \beta \map \nu A$

is well-defined for each $A \in \Sigma$.

Then:

$\xi = \alpha \mu + \beta \nu$ is a signed measure.

Proof

We verify both of the conditions for a signed measure.

Proof of $(1)$

We have:

\(\ds \map \xi \O\)

\(=\)

\(\ds \alpha \map \mu \O + \beta \map \nu \O\)

\(\ds \)

\(=\)

\(\ds \alpha \times 0 + \beta \times 0\)

since $\mu$ and $\nu$ are signed measures we have $\map \mu \O = \map \nu \O = 0$

\(\ds \)

\(=\)

\(\ds 0\)

verifying $(1)$ for $\xi$.

$\Box$

Proof of $(2)$

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

Then:

\(\ds \map \xi {\bigcup_{n \mathop = 1}^\infty D_n}\)

\(=\)

\(\ds \alpha \map \mu {\bigcup_{n \mathop = 1}^\infty D_n} + \beta \map \nu {\bigcup_{n \mathop = 1}^\infty D_n}\)

\(\ds \)

\(=\)

\(\ds \alpha \sum_{n \mathop = 1}^\infty \map \mu {D_n} + \beta \sum_{n \mathop = 1}^\infty \map \nu {D_n}\)

since $\mu$ and $\nu$ are countably additive

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \paren {\alpha \map \mu {D_n} + \beta \map \nu {D_n} }\)

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \map \xi {D_n}\)

verifying $(2)$ for $\xi$.

$\Box$

So $\xi$ is a signed measure.

$\blacksquare$