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:


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

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: