Integral of Integrable Function is Homogeneous

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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:

\(\ds \int \paren {\lambda f}^+ \rd \mu\) \(=\) \(\ds \int \lambda f^+ \rd \mu\)
\(\ds \) \(=\) \(\ds \lambda \int f^+ \rd \mu\) Integral of Positive Measurable Function is Positive Homogeneous
\(\ds \) \(<\) \(\ds \infty\) since $f$ is $\mu$-integrable

Similarly:

\(\ds \int \paren {\lambda f}^- \rd \mu\) \(=\) \(\ds \int \lambda f^- \rd \mu\)
\(\ds \) \(=\) \(\ds \lambda \int f^- \rd \mu\) Integral of Positive Measurable Function is Positive Homogeneous
\(\ds \) \(<\) \(\ds \infty\) since $f$ is $\mu$-integrable

So:

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

We then have:

\(\ds \int \lambda f \rd \mu\) \(=\) \(\ds \int \paren {\lambda f}^+ \rd \mu - \int \paren {\lambda f}^- \rd \mu\) Definition of Integral of Integrable Function
\(\ds \) \(=\) \(\ds \lambda \int f^+ \rd \mu - \lambda \int f^- \rd \mu\) Positive Part of Multiple of Function, Negative Part of Multiple of Function
\(\ds \) \(=\) \(\ds \lambda \paren {\int f^+ \rd \mu - \int f^- \rd \mu}\)
\(\ds \) \(=\) \(\ds \lambda \int f \rd \mu\) Definition of Integral of Integrable Function


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:

\(\ds \int \paren {\lambda f}^+ \rd \mu\) \(=\) \(\ds \int \paren {-\lambda f^-} \rd \mu\)
\(\ds \) \(=\) \(\ds -\lambda \int f^- \rd \mu\) Integral of Positive Measurable Function is Positive Homogeneous, since $-\lambda \ge 0$
\(\ds \) \(<\) \(\ds \infty\) since $f$ is $\mu$-integrable

Similarly:

\(\ds \int \paren {\lambda f}^- \rd \mu\) \(=\) \(\ds \int \paren {-\lambda f^+} \rd \mu\)
\(\ds \) \(=\) \(\ds -\lambda \int f^+ \rd \mu\) Integral of Positive Measurable Function is Positive Homogeneous
\(\ds \) \(<\) \(\ds \infty\) since $f$ is $\mu$-integrable

So:

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

We then have:

\(\ds \int \lambda f \rd \mu\) \(=\) \(\ds \int \paren {\lambda f}^+ \rd \mu - \int \paren {\lambda f}^- \rd \mu\) Definition of Integral of Integrable Function
\(\ds \) \(=\) \(\ds \int \paren {-\lambda f^-} \rd \mu - \int \paren {-\lambda f^+} \rd \mu\) Positive Part of Multiple of Function, Negative Part of Multiple of Function
\(\ds \) \(=\) \(\ds \lambda \int f^+ \rd \mu - \lambda \int f^- \rd \mu\) Integral of Positive Measurable Function is Positive Homogeneous
\(\ds \) \(=\) \(\ds \lambda \paren {\int f^+ \rd \mu - \int f^- \rd \mu}\)
\(\ds \) \(=\) \(\ds \lambda \int f \rd \mu\) Definition of Integral of Integrable Function

$\blacksquare$


Sources

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