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:

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.