Set of Points for which Measurable Function is Real-Valued is Measurable
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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.
$\blacksquare$