Measure Vector Space
Why
If both signed measures are finite, then their difference is always well-defined. Is the difference a finite signed measure?
Preliminary result
Let $(X, \mathcal{A} )$ be a measurable space.
Let $\mu $ and $\nu $ be finite signed
measures.
Let $R$ denote the real numbers.
Then $(\alpha \mu )(\varnothing) = \alpha \cdot
\mu (\varnothing) = \alpha \cdot 0 = 0$.
Also for $\seq{A} \subset \mathcal{A} $
disjoint,
\[
\begin{aligned}
(\alpha \mu )(\cup A_n) &= \alpha \mu (\cup A_n) = \alpha
\sum_{n = 1}^{\infty} \mu (A_n) \\
&= \sum_{n = 1}^{\infty} \alpha \mu (A_n) = (\alpha \mu )(A_n)
\end{aligned}
\]
Similarly, $(\mu + \nu )(\varnothing) =
\mu (\varnothing) + \nu (\varnothing) = 0$.
And, for $\seq{A} \subset \mathcal{A} $
disjoint,
\[
\begin{aligned}
(\mu + \nu )(\cup A_n) &= \mu (\cup A_n) + \nu (\cup A_n)
= \sum_{n = 1}^{\infty} \mu (A_n) +
\sum_{n = 1}^{\infty} \nu (A_n) \\
&= \sum_{n = 1}^{\infty} \mu (A_n) + \nu (A_n)
= \sum_{n = 1}^{\infty} (\mu + \nu )(A_n)
\end{aligned}
\]
Main result
Notation
We denote the vector space of signed measures on measurable space $(X, \mathcal{A} )$ by by $M(X, \mathcal{A} , \R )$.