Measurable Function is Integrable iff A.E. Equal to Real-Valued Integrable Function
Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Then $f$ is $\mu$-integrable if and only if:
- there exists a $\mu$-integrable function $g : X \to \R$ such that $g = f$ $\mu$-almost everywhere.
Proof
Sufficient Condition
Suppose that:
- there exists a $\mu$-integrable function $g : X \to \R$ such that $g = f$ $\mu$-almost everywhere.
Then, from A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have:
- $f$ is $\mu$-integrable.
$\Box$
Necessary Condition
Suppose that $f$ is $\mu$-integrable.
From Integrable Function is A.E. Real-Valued, we have:
- $\map f x \in \R$ for $\mu$-almost all $x \in X$.
That is, there exists a $\mu$-null set $N \subseteq X$ such that whenever:
- $\size {\map f x} = \infty$
we have $x \in N$.
Define a function $g : X \to \R$ by:
- $\map g x = \map f x \map {\chi_{X \setminus N} } x$
for each $x \in X$.
Since $N \in \Sigma$, we have:
- $X \setminus N \in \Sigma$
From Characteristic Function Measurable iff Set Measurable, we have:
- $\chi_{X \setminus N}$ is $\Sigma$-measurable.
From Pointwise Product of Measurable Functions is Measurable, we have:
- $g$ is $\Sigma$-measurable.
Note that whenever $x \in X$ has:
- $\map f x \map {\chi_{X \setminus N} } x \ne \map f x$
we have:
- $\map {\chi_{X \setminus N} } x = 0$
That is, from the definition of set difference:
- $x \in N$
So:
- $g = f$ $\mu$-almost everywhere.
From A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have:
- $g$ is $\mu$-integrable.
So:
- there exists a $\mu$-integrable function $g : X \to \R$ such that $g = f$ $\mu$-almost everywhere.
$\blacksquare$
Sources
- 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $2.3$: Integral