Integral of Bounded Measurable Function with respect to Finite Signed Measure is Well-Defined

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

Let $f : X \to \R$ be a bounded $\Sigma$-measurable function.

Let $\mu$ be a finite signed measure on $\struct {X, \Sigma}$.

Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.

Then the $\mu$-integral of $f$ defined by:


 * $\ds \int f \rd \mu = \int f \rd \mu^+ - \int f \rd \mu^-$

is well-defined.

Proof
We show that $f$ is $\mu^+$-integrable and $\mu^-$-integrable.

We will then have:


 * $\ds -\infty < \int f \rd \mu^+ < \infty$

and:


 * $\ds -\infty < \int f \rd \mu^- < \infty$

So that:


 * $\ds \int f \rd \mu^+ - \int f \rd \mu^-$

is well-defined.

Since $f$ is bounded, there exists $M > 0$ such that:


 * $\size {\map f x} \le M$

for each $x \in X$.

From Jordan Decomposition of Finite Signed Measure, we have:


 * $\mu^+$ and $\mu^-$ are finite.

That is:


 * $\map {\mu^+} X < \infty$

and:


 * $\map {\mu^-} X < \infty$

We therefore have:

so:

So:


 * $\size f$ is $\mu^+$-integrable

and:


 * $\size f$ is $\mu^-$-integrable.

From Characterization of Integrable Functions, we have:


 * $f$ is $\mu^+$-integrable and $\mu^-$-integrable

as required.