Measurable Function is Integrable iff A.E. Equal to Real-Valued Integrable Function

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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

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