# Set of Points for which Measurable Function is Real-Valued is Measurable

## Theorem

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

Let $f: X \to \overline \R$ be a $\Sigma$-measurable.

Then:

$\set {x \in X : \map f x \in \R}$ is $\Sigma$-measurable.

### Corollary

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

## Proof

Since $f$ is $\Sigma$-measurable, we have that:

for all $n \in \N$ the set $\set {x \in X : \map f x \le n}$ is $\Sigma$-measurable

and:

for all $n \in \N$ the set $\set {x \in X : -n \le \map f x}$ is $\Sigma$-measurable.

From $\sigma$-Algebra Closed under Countable Intersection, we have:

$\set {x \in X : -n \le \map f x \le n} = \set {x \in X : \map f x \le n} \cap \set {x \in X : -n \le \map f x}$ is $\Sigma$-measurable.

Since $\sigma$-algebras are closed under countable union, we also have:

$\ds \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$ is $\Sigma$-measurable.

We will finally show that:

$\ds \set {x \in X : \map f x \in \R} = \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$

which gives the claim.

If $x \in X$ has $\map f x \in \R$, then we have:

$-\paren {\floor {\size {\map f x} } + 1} \le \map f x \le \floor {\size {\map f x} } + 1$

with:

$\floor {\size {\map f x} } + 1 \in \N$

So:

$\ds x \in \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$

Now, let:

$\ds x \in \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$

Then:

$-n \le \map f x \le n$

for some $n \in \N$, so certainly:

$\map f x \in \R$

So we have:

$\ds \set {x \in X : \map f x \in \R} = \bigcup_{n \mathop = 1}^\infty \set {x \in X : -n \le \map f x \le n}$

and hence the claim.

$\blacksquare$