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:


 * $(1): \quad f \in \mathcal{L}_{\overline{\R}} \left({\mu}\right)$, i.e., $f$ 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 $due to 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


 * $(4) \implies (1)\quad$ will be detailed as follows:


 * Let $f \in \mathcal{M}_{\overline{\R}}$ and $g$ according to $(4)$


 * $f = f^+ - f^-$ according positive and negative parts


 * $f^+$ and $f^-$ are positive and measurable


 * Let $f^0$ stand for any of $f^+$ or $f^-$


 * $f^0 \le \left\vert{f}\right\vert \le g$ because $\left\vert{f}\right\vert = f^+ +\, f^-$


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


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


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


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


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


 * Because integrals for $f^+$ and $f^-$ both are finite, $f$ is $\mu$-integrable according to definition