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 :


 * there exists a $\mu$-integrable function $g : X \to \R$ such that $g = f$ $\mu$-almost everywhere.

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.

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.