Integrable Function is A.E. Real-Valued

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

Let $f: X \to \overline \R$ be a $\mu$-integrable function.

Then $\map f x \in \R$ for almost all $x \in X$.

Proof
We have that $\map f x \not \in \R$ if:


 * $\size {\map f x} = +\infty$

This is the case :


 * $\size {\map f x} \ge n$

for each $n \in \N$.

So, we can write:


 * $\ds \set {x \in X : \size {\map f x} = +\infty} = \bigcap_{n \mathop = 1}^\infty \set {x \in X : \size {\map f x} \ge n}$

We have that:


 * $\ds \set {x \in X : \size {\map f x} \ge n} \in \Sigma$

for each $n \in \N$, by Characterization of Measurable Functions.

Since $\Sigma$ is a $\sigma$-algebra, we have that:


 * $\ds \bigcap_{n \mathop = 1}^\infty \set {x \in X : \size {\map f x} \ge n} \in \Sigma$

as the countable intersection of sets in $\Sigma$.

So:


 * $\set {x \in X : \size {\map f x} = +\infty} \in \Sigma$

We now aim to show that this set is a null set.

Now note that we have:


 * $\set {x \in X : \size {\map f x} = +\infty} \subseteq \set {x \in X : \size {\map f x} \ge n}$

From Measure is Monotone, we therefore have:


 * $\ds \map \mu {\set {x \in X : \size {\map f x} = +\infty} } \le \map \mu {\set {x \in X : \size {\map f x} \ge n} }$

From Markov's Inequality, we have:


 * $\ds \map \mu {\set {x \in X : \size {\map f x} \ge n} } \le \frac 1 n \int \size f \rd \mu$

So:


 * $\ds 0 \le \map \mu {\set {x \in X : \size {\map f x} = +\infty} } \le \frac 1 n \int \size f \rd \mu$

for each $n \in \N$.

From Lower and Upper Bounds for Sequences, taking $n \to \infty$, we obtain:


 * $\ds \map \mu {\set {x \in X : \size {\map f x} = +\infty} } = 0$

That is:


 * $\ds \map \mu {X \setminus \set {x \in X : \map f x \in \R} } = 0$

So:


 * $\map f x \in \R$ for almost all $x \in X$.