Integral of Integrable Function is Additive

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:


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

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:

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


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

Then:

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:

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: