$f=g$ almost everywhere $\Rightarrow |f|=|g|$ almost everywhere? Suppose $(X, \mathcal{M}, \mu)$ is a measure space. Assume $f: X\to\overline{\mathbb{R}}$ and $g=X\to\overline{\mathbb{R}}$ are measurable maps. Here $\overline{\mathbb{R}}$ denotes the set of extended real numbers. My question is:

If $f=g$ almost everywhere, does it follow that $|f|=|g|$ almost
  everywhere?

I know the answer is "Yes" if $X$ is a complete measure space: If $f=g$ a.e. then $E=\{x\in X: f(x)\neq g(x)\}$ is a null set, i.e. $\mu(E)=0$. It is clear that
$$
F=\{x\in X : |f(x)|\neq |g(x)|\}\subseteq E
$$
Since $X$ is complete, all subsets of null sets are in $\mathcal{M}$, and so $\mu(F)=0$, and $|f|=|g|$ a.e. 
What happens when $X$ is not complete?
Thanks for your time :)
 A: Since $f,g$ are measurable, so are $f-g$ and $f+g$. Let
$$
E=\{x\in X:\ f(x)=g(x)\}, \ \ E'=\{x\in X:\ f(x)=-g(x)\}.
$$
If $F$ is as in the question, the set where $|f|\neq|g|$, then we have $F^{c}=E\cup E'$. And then $F$ is measurable, because both $E$ and $E'$ are:
$$
E=(f-g)^{-1}(\{0\}),\ \ E'=(f+g)^{-1}(\{0\}).
$$
A: Note that since $f$ and $g$ are measurable, then so are $|f|$ and $|g|$ by continuity of $|\cdot|$, and hence, $h=|f|-|g|$ is measurable. Noting that $$F=\left(h^{-1}\bigl(\{0\}\bigr)\right)^c,$$ we readily have $F\in\mathcal M$. By nonnegativity and monotonicity of measure, since $F\subseteq E$ and $\mu(E)=0,$ then $\mu(F)=0,$ and so $|f|=|g|$ a.e.
There is no need for $(X,\mathcal M,\mu)$ to be a complete measure space.
A: If f=g a.e. that means that h=f-g=0 a.e. which clearly implies |h(x)|=0 a.e. i.e. |f(x)-g(x)|=0 almost everywhere! But ||f(x)-|g(x)||$\leqslant$ |f(x)-g(x)| which is zero on a set of full measure and hence ||f(x)|-|g(x)||=0 on this set of full measure so |f(x)|=|g(x)| a.e
