A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra

By [L1], choose simple $\overline{{\mathcal{A}}_0}$-measurable functions [L1, choose] $s_k:X\to \mathbb{R}$ with $\mathrm{\mid }{s}_k\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$ and $s_k(x)\to f(x)$ for every $x\in X$. Write the canonical representation of $s_k$ as $s_k={\sum }_{j=1}^{m_k}{c}_{k,j}{\mathbf{1}}_{E_{k,j}},$ where the $E_{k,j}$ are pairwise disjoint completed measurable level sets.

For each pair $(k,j)$, apply [L2] to $E_{k,j}$ and choose [L2, L3, choose] $A_{k,j}\in {\mathcal{A}}_0$ together with a completed null set $N_{k,j}$ such that $E_{k,j}=A_{k,j}\cup M_{k,j}$ with $M_{k,j}\subseteq N_{k,j}$. Because $A_{k,j}\subseteq E_{k,j}$, the sets $A_{k,j}$ remain pairwise disjoint. Define $t_k:={\sum }_{j=1}^{m_k}{c}_{k,j}{\mathbf{1}}_{A_{k,j}}\mathrm{.}$ Then each $t_k$ is ${\mathcal{A}}_0$-measurable and simple. Let $N:={\bigcup }_{k,j}{N}_{k,j}$. By [L3], $N$ is a completed measurable null set, and for every $x\mathrm{\notin }N$ one has $t_k(x)=s_k(x)$ for all $k$.

Define [step 1.1, step 1.2, L4] $g:={\mathrm{lim\hspace{0.17em}sup}}_{k\to \mathrm{\infty }}{t}_k\mathrm{.}$ By [L4], the function $g$ is ${\mathcal{A}}_0$-measurable. If $x\mathrm{\notin }N$, then step 1.2 gives $t_k(x)=s_k(x)$ for every $k$, and step 1.1 gives $s_k(x)\to f(x)$, so $g(x)={\mathrm{lim\hspace{0.17em}sup}}_k{t}_k(x)={\lim }_k{s}_k(x)=f(x)$. Hence $g=f$ on $X\setminus N$.

The null set $N$ is measurable in the completion by [L3], so step 2.1 says [step 2.1, L3] exactly that $f=g$ almost everywhere. Since $g$ is ${\mathcal{A}}_0$-measurable, it is the required base-measurable representative.