Integral of Integrable Function is Homogeneous

Theorem

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

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

Let $\lambda \in \R$.

Let $\lambda f$ be the pointwise $\lambda$-multiple of $f$.

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

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

Complex Function

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

First suppose that $\lambda \ge 0$.

From Positive Part of Multiple of Function, we have:

$\paren {\lambda f}^+ = \lambda f^+$

From Negative Part of Multiple of Function, we have:

$\paren {\lambda f}^- = \lambda f^-$

From Function Measurable iff Positive and Negative Parts Measurable, we have:

$f^-$ and $f^+$ are $\Sigma$-measurable.

From Pointwise Scalar Multiple of Measurable Function is Measurable, we have:

$\lambda f^-$, $\lambda f^+$ and $\lambda f$ are $\Sigma$-measurable.

Then, we have:

Similarly:

So:

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

We then have:

Now suppose that $\lambda < 0$.

From Positive Part of Multiple of Function, we have:

$\paren {\lambda f}^+ = -\lambda f^-$

From Negative Part of Multiple of Function, we have:

$\paren {\lambda f}^- = -\lambda f^+$

From Function Measurable iff Positive and Negative Parts Measurable, we have:

$f^-$ and $f^+$ are $\Sigma$-measurable.

From Pointwise Scalar Multiple of Measurable Function is Measurable, we have:

$\lambda f^-$, $\lambda f^+$ and $\lambda f$ are $\Sigma$-measurable.

Then we have:

Similarly:

So:

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

We then have:

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $10.4 \ \text{(i)}$