# 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.

$\blacksquare$

.