Characterization of Integrable Functions

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Then the following are equivalent:

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



Proof

We prove the whole cycle of implications:

$(1) \implies (2) \quad$ by definition of $(1)$
$(2) \implies (3)\quad$ because $\size f = f^+ + f^-$ and Integral of Positive Measurable Function is Additive
$(3) \implies (4)\quad$ because $g:= \size f$ exists


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

Let $f \in \MM_{\overline \R}$ 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:

$\size f = f^+ + f^-$

Therefore:

$f^0 \le \size f \le g$

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

Let $\EE^+$ and $\map {I_\mu} h$ be defined as in Integral of Positive Measurable Function.

Then:

$\forall h \in \EE^+$: $h \le f^0 \implies h \le g$

Hence:

$\set {h: h \le f^0, h \in \EE^+} \subseteq \set {h: h \le g, h \in \EE^+}$
$\ds \int f^0 \rd \mu := \sup \set {\map {I_\mu} h: h \le f^0, h \in \EE^+} \le \sup \set {\map {I_\mu} h: h \le g, h \in \EE^+} < \infty$

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

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

$\blacksquare$



.

Sources