Integral of Integrable Function is Homogeneous/Complex Function

Theorem

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

Let $\struct {\C, \map \BB \C}$ be the complex numbers made into a measurable space with its Borel $\sigma$-algebra.

Let $f : X \to \C$ be a $\mu$-integrable function.

Let $\lambda \in \C$.

Then $\lambda f$ is $\mu$-integrable and:

$\ds \int \lambda f \rd \mu = \lambda \int f \rd \mu$

Proof

We have:

From Real Part of Complex Product, we have:

$\map \Re {\lambda f} = \map \Re \lambda \map \Re f - \map \Im \lambda \map \Im f$

From Imaginary Part of Complex Product, we have:

$\map \Im {\lambda f} = \map \Im \lambda \map \Re f + \map \Re \lambda \map \Im f$

Since $f$ is $\mu$-integrable, $\map \Re f$ and $\map \Im f$ are $\mu$-integrable by definition.

From Integral of Integrable Function is Additive and Integral of Integrable Function is Homogeneous, $\map \Re {\lambda f}$ and $\map \Im {\lambda f}$ are both $\mu$-integrable and we have:

$\ds \int \map \Re {\lambda f} \rd \mu = \map \Re \lambda \int \map \Re f \rd \mu - \map \Im \lambda \int \map \Im f \rd \mu$

and:

$\ds \int \map \Im {\lambda f} \rd \mu = \map \Im \lambda \int \map \Re f \rd \mu + \map \Re \lambda \int \map \Im f \rd \mu$

Then $\lambda f$ is $\mu$-integrable.

We then have:

$\blacksquare$