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
From Set of Points for which Measurable Function is Real-Valued is Measurable: Corollary, we have that:


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

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 $x \in X$.