Integral of Integrable Function is Additive

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Theorem

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

Let $f, g: X \to \R$ be $\mu$-integrable functions.


Then $f + g$ is $\mu$-integrable, with:

$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$


Corollary 1

Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.

Suppose that the pointwise sum $f + g$ is well-defined.


Then $f + g$ is $\mu$-integrable, with:

$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$


Corollary 2

Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.

Suppose that the pointwise difference $f - g$ is well-defined.


Then $f - g$ is $\mu$-integrable, with:

$\ds \int \paren {f - g} \rd \mu = \int f \rd \mu - \int g \rd \mu$


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


Then $f + g$ is $\mu$-integrable and:

$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$


Proof

From Pointwise Sum of Measurable Functions is Measurable:

$f + g$ is $\Sigma$-measurable.

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

$\paren {f + g}^+$ and $\paren {f + g}^-$ are $\Sigma$-measurable.

Since $f$ and $g$ are $\mu$-integrable, we have that $f$ and $g$ are also $\Sigma$-measurable.

Then from Function Measurable iff Positive and Negative Parts Measurable we have that:

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

We now want to show that:

$\ds \int \paren {f + g}^+ \rd \mu < \infty$

and:

$\ds \int \paren {f + g}^- \rd \mu < \infty$


From Bound for Positive Part of Pointwise Sum of Functions, we have:

$\paren {f + g}^+ \le f^+ + g^+$

Then:

\(\ds \int \paren {f + g}^+ \rd \mu\) \(\le\) \(\ds \int \paren {f^+ + g^+} \rd \mu\)
\(\ds \) \(=\) \(\ds \int f^+ \rd \mu + \int g^+ \rd \mu\) Integral of Positive Measurable Function is Additive
\(\ds \) \(<\) \(\ds \infty\) since $f$ and $g$ are $\mu$-integrable, both integrals are finite


From Bound for Negative Part of Pointwise Sum of Functions, we have:

$\paren {f + g}^- \le f^- + g^-$

Then:

\(\ds \int \paren {f + g}^- \rd \mu\) \(\le\) \(\ds \int \paren {f^- + g^-} \rd \mu\)
\(\ds \) \(=\) \(\ds \int f^- \rd \mu + \int g^- \rd \mu\) Integral of Positive Measurable Function is Additive
\(\ds \) \(<\) \(\ds \infty\) since $f$ and $g$ are $\mu$-integrable, both integrals are finite


Now, we will show that:

$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$

From Difference of Positive and Negative Parts, we have:

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

On the other hand, we have:

\(\ds f + g\) \(=\) \(\ds \paren {f^+ - f^-} + \paren {g^+ - g^-}\) Difference of Positive and Negative Parts
\(\ds \) \(=\) \(\ds \paren {f^+ + g^+} - \paren {f^- + g^-}\)

That is:

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

From Integral of Positive Measurable Function is Additive, we have:

$f^+ + g^+$ and $f^- + g^-$ are $\mu$-integrable.

From Integral of Integrable Function is Additive: Lemma, we then have:

$\ds \int \paren {f^+ + g^+} \rd \mu - \int \paren {f^- + g^-} \rd \mu = \int \paren {f + g}^+ \rd \mu - \int \paren {f + g}^- \rd \mu$

From Integral of Positive Measurable Function is Additive, we have:

$\ds \int \paren {f^+ + g^+} \rd \mu = \int f^+ \rd \mu + \int g^+ \rd \mu$

and:

$\ds \int \paren {f^- + g^-} \rd \mu = \int f^- \rd \mu + \int g^- \rd \mu$

Hence:

\(\ds \int \paren {f + g} \rd \mu\) \(=\) \(\ds \int \paren {f + g}^+ \rd \mu - \int \paren {f + g}^- \rd \mu\) Definition of Integral of Integrable Function
\(\ds \) \(=\) \(\ds \int f^+ \rd \mu + \int g^+ \rd \mu - \paren {\int f^- \rd \mu + \int g^- \rd \mu}\)
\(\ds \) \(=\) \(\ds \paren {\int f^+ \rd \mu - \int f^- \rd \mu} + \paren {\int g^+ \rd \mu - \int g^- \rd \mu}\)
\(\ds \) \(=\) \(\ds \int f \rd \mu + \int g \rd \mu\) Definition of Integral of Integrable Function

$\blacksquare$


Sources

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