Characterization of Integrable Functions

Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $f: X \to \overline \R, f \in \mathcal M_{\overline \R}$ be a $\Sigma$-measurable function.

Then the following are equivalent:


 * $(1): \quad f \in \mathcal L_{\overline \R} \left({\mu}\right)$, that is, $f$ is $\mu$-integrable.
 * $(2): \quad$ The positive and negative parts $f^+$ and $f^-$ are $\mu$-integrable.
 * $(3): \quad$ The absolute value $\left\vert{f}\right\vert$ of $f$ is $\mu$-integrable.
 * $(4): \quad$ There exists an $\mu$-integrable function $g: X \to \overline \R$ such that $\left\vert{f}\right\vert \le g$ pointwise.

Proof
We prove the whole cycle of implications:


 * $(1) \implies (2) \quad$ by definition of $(1)$


 * $(2) \implies (3)\quad$ because $\left\vert{f}\right\vert = f^+ +\, f^-$ and Integral of Positive Measurable Function is Additive


 * $(3) \implies (4)\quad$ because $g:= \left\vert{f}\right\vert$ exists

It remains to demonstrate $(4) \implies (1)$.

Let $f \in \mathcal M_{\overline \R}$ and $g$ according to $(4)$.

Then:
 * $f = f^+ - f^-$

where $f^+$ is the positive and $f^-$ is the negative part of $f$.

We have that $f^+$ and $f^-$ are positive and measurable.

Let $f^0$ stand for either $f^+$ or $f^-$.

We have that:
 * $\left\vert{f}\right\vert = f^+ + f^-$

Therefore:
 * $f^0 \le \left\vert{f}\right\vert \le g$

It is to be shown that the Integral of Positive Measurable Function of $f^0$ exists and is finite.

Let $\mathcal E^+$ and $I_\mu \left({h}\right)$ be defined as in Integral of Positive Measurable Function.

Then:
 * $\forall h \in \mathcal E^+$: $h \le f^0 \implies h \le g$

Hence:


 * $\left\{{h: h \le f^0, h \in \mathcal E^+}\right\} \subseteq \left\{ {h: h \le g, h \in \mathcal E^+}\right\}$


 * $\displaystyle \int f^0 \, \mathrm d \mu := \sup \left\{ {I_\mu \left({h}\right): h \le f^0, h \in \mathcal E^+}\right\} \le \sup \left\{{I_\mu \left({h}\right): h \le g, h \in \mathcal E^+}\right\} \lt \infty$

We have that the integrals for $f^+$ and $f^-$ both are finite

Therefore $f$ is $\mu$-integrable according to definition.

.