# Showing Functions are Measurable on Complete Measure Spaces

Let $(X,\mathcal A,\mu)$ be a complete measure space and $f: X\to \mathbb R$ is a measurable function. If $g: X\to \mathbb R$ is a function such that $Z:= \{x\in X\,| f(x) \not=g(x)\}$ has zero measure. Show that g is also measurable and give an example to show why $(X,\mathcal A,\mu)$ must be complete.

I am working on this problem but am having lots of difficulties.

Firstly we know that because $(X,\mathcal A,\mu)$ is a complete measure space then all subsets of $Z$ are also measurable.

I have tried two different approaches, the first was attempting to define $E :=\{x \in X\, | f(x) = g(x)\}$ for $E \subseteq Z$ and somehow relate this set E which we know is measurable to $g(x)$ but I do not believe this is the correct approach as we do not know anything about when $f(x) = g(x)$.

My second approach was that to suppose that $g(x)$ is not measurable and find a contradiction by using the following definition of a measurable function:

If $(X,\mathcal A)$ is a measurable space, a measurable function $h: X\to \mathbb R$ is a function $h$ such that $h^{-1}(\text{measurable }E \subseteq \mathbb R) \in \mathcal A$

From here I believe assuming that $g$ is not measurable and then trying to show the existence of such an $E \subseteq Z$ which $g$ is defined on.

Thank you, any questions, comments or answers or improvements are greatly appreciated.

Since sets of the form $(-\infty,\alpha)$ generate the Borel $\sigma$-algebra on $\mathbb{R}$ it suffices to check that $\{g<\alpha\}$ is measurable for all $\alpha\in\mathbb{R}$.
We can write $$\{g<\alpha\}=\left(\{f<\alpha\}\cup\left(\{g<\alpha\}\setminus\{f<\alpha\}\right)\right)\setminus\left(\{f<\alpha\}\setminus\{g<\alpha\}\right)$$ Now note that $$\{g<\alpha\}\setminus\{f<\alpha\}\subseteq\{f\neq g\},\qquad \{f<\alpha\}\setminus\{g<\alpha\}\subseteq\{f\neq g\}.$$ By the completeness of $\mu$, these last two sets are measurable, and since $f$ is measurable $\{f<\alpha\}$ is measurable and so $\{g<\alpha\}$ is measurable as seen by the above partitioning.
• Yup, that is exactly why we only need to consider $\{g<\alpha\}$. Note that this was in no way necessary... If instead we considered any measurable set $E\subseteq\mathbb{R}$ then we can write $g^{-1}(E)=(f^{-1}(E)\cup(g^{-1}(E)\setminus f^{-1}(E)))\setminus (f^{-1}(E)\setminus g^{-1}(E))$ and the rest of the proof still goes through since $g^{-1}(E)\setminus f^{-1}(E)\subseteq\{f\neq g\}$, etc. – yousuf soliman Aug 20 '17 at 8:27