Function A.E. Equal to Measurable Function in Complete Measure Space is Measurable
Theorem
Let $\struct {X, \Sigma, \mu}$ be a complete measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $g : X \to \overline \R$ be a function such that:
- $f = g$ $\mu$-almost everywhere.
Then $g$ is $\Sigma$-measurable.
Proof
We aim to show that:
- $\set {x \in X : \map g x \le \alpha} \in \Sigma$
for each $\alpha \in \R$.
Let $\alpha \in \R$.
Since $f = g$ $\mu$-almost everywhere there exists a $\mu$-null set such that:
- whenever $x \in X$ has $\map f x \ne \map g x$, we have $x \in N$.
We have:
\(\ds \set {x \in X : \map g x \le \alpha}\) | \(=\) | \(\ds \set {x \in X : \map g x \le \alpha} \cap \paren {N \cup \paren {X \setminus N} }\) | Intersection with Subset is Subset, Union with Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\set {x \in X : \map g x \le \alpha} \cap N} \cup \paren {\set {x \in X : \map g x \le \alpha} \cap \paren {X \setminus N} }\) | Union Distributes over Intersection |
For $x \in X \setminus N$, we have:
- $\map g x = \map f x$
so:
- $\set {x \in X \setminus N : \map g x \le \alpha} \cap \paren {X \setminus N} = \set {x \in X \setminus N : \map f x \le \alpha} \cap \paren {X \setminus N}$
Since $f$ is $\Sigma$-measurable, we have:
- $\set {x \in X : \map f x \le \alpha} \in \Sigma$
Since $\sigma$-algebras are closed under complement, we have:
- $X \setminus N \in \Sigma$
From Sigma-Algebra Closed under Countable Intersection, we therefore have:
- $\set {x \in X \setminus N : \map f x \le \alpha} \cap \paren {X \setminus N} \in \Sigma$
Now, from Intersection is Subset, we have:
- $\set {x \in X : \map g x \le \alpha} \cap N \subseteq N$
Since $N$ is $\mu$-null, we have:
- $\set {x \in X : \map g x \le \alpha}$ is $\mu$-null
since $\struct {X, \Sigma, \mu}$ is complete.
In particular:
- $\set {x \in X : \map g x \le \alpha} \cap N \in \Sigma$
So, since $\sigma$-algebras are closed under countable union, we have:
- $\paren {\set {x \in X : \map g x \le \alpha} \cap N} \cup \paren {\set {x \in X : \map g x \le \alpha} \cap \paren {X \setminus N} } \in \Sigma$
that is:
- $\set {x \in X : \map g x \le \alpha} \in \Sigma$
Since $\alpha \in \R$ was arbitrary, we have:
- $g$ is $\Sigma$-measurable.
$\blacksquare$