Measurability of function on incomplete measurable spaces.

Question

I have this problem:

Let $(X,\mathcal{F},\mu)$ a NON complete measurable space. If f is measurable in X, prove that there exist a non measurable function g: X $\rightarrow\bar{\mathbb{R}}$ such that f(x) = g(x) almost everywhere (w.r.t $\mu$).

My attempt:

Since $(X,\mathcal{F},\mu)$ is not complete, there will exist a set B$\subset$C where C is a measurable set with $\mu{(C)}=0$

Now I define the function:

$ g(x)=\begin{cases} +\infty & \mbox{in }\mbox{B} \\ f(x) & \mbox{in } \mbox{ X\C}\\0 & \mbox{in } \mbox{C\B} \end{cases} $

This function is not measurable because, for instance, if we choose $\alpha>0$

$\{g>\alpha\}$ = $\{f>\alpha\}\cup\{g>\alpha\}$. The second set is B and is not measurable.

Am I right or there are better way to answer?

Thanks!

Answer 1

Just use the indicator function of your non-measurable set. If $(X, \mathcal{F}, \mu)$ is not complete then there's a null set $C$ and non-measurable $B \subset C$. Since $B$ is non-measurable, so is $\mathbf{1}_B$. Therefore, $g = f + \mathbf{1}_B$ is non-measurable (if $g$ were measurable then $\mathbf{1}_B = g-f$ would be measurable as well; a contradiction).

But $g(x) = f(x)$ for all $x \in X \setminus C$, which is to say almost everywhere.