Pointwise Sum of Integrable Functions is Integrable Function

Theorem

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

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

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

Then $f + g$ is also a $\mu$-integrable function.

That is, the space of $\mu$-integrable functions $\LL^1_{\overline \R}$ is closed under pointwise addition.

Proof

We are given $f, g: X \to \overline \R$ are $\mu$-integrable functions:

$\ds \int f^+ \rd \mu$

$<$

$\ds +\infty$

$\ds \int f^- \rd \mu$

$<$

$\ds +\infty$

where $f^+$ and $f^-$ are the positive and negative parts of $f$ respectively.

Also:

$\ds \int g^+ \rd \mu$

$<$

$\ds +\infty$

$\ds \int g^- \rd \mu$

$<$

$\ds +\infty$

We are given that pointwise sum $f + g$ is well-defined.

So by Bound for Positive Part of Pointwise Sum of Functions, there exists no $x \in X$ such that:

$\set {\map f x, \map g x} = \set {\infty, -\infty}$

and:

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

By Bound for Negative Part of Pointwise Sum of Functions, there exists no $x \in X$ such that:

$\set {\map f x, \map g x} = \set {\infty, -\infty}$

and:

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

This needs considerable tedious hard slog to complete it.
In particular: Link to below conclusion should be included
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Therefore:

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

$<$

$\ds +\infty$

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

$<$

$\ds +\infty$

Hence $f + g$ is also a $\mu$-integrable function.

$\blacksquare$

Sources

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