= Solution
The pointwise limit $f$ is a <measurable function>, since it can be expressed as $\sup_n\inf_{m\ge n}f_m$. Thus each <set> $E_n^{(k)}$ is a countable intersection of measurable <sets> within $A$. Deleting the condition at index $m=n$ weakens the requirements, whereas replacing $1/k$ by $1/(k+1)$ strengthens them. Hence
$$
\boxed{E_n^{(k)}\subseteq E_{n+1}^{(k)},\qquad E_n^{(k+1)}\subseteq E_n^{(k)}}.
$$
These are inclusions, not necessarily strict inclusions. The displayed <sets> describe eventual uniform error tolerances at an individual point.
Fix $k\ge1$ and $x\in A$. By <pointwise convergence>, there is an $N=N(x,k)$ such that $|f_m(x)-f(x)|\le1/k$ for every $m\ge N$. This is exactly $x\in E_N^{(k)}$, proving
$$
\boxed{A=\bigcup_{n\ge1}E_n^{(k)}}.
$$
Together with monotonicity in $n$ and finite measure of $A$, <continuity from below of a measure> gives $\lambda(E_n^{(k)})\uparrow\lambda(A)$. Consequently
$$
\lambda(A\setminus E_n^{(k)})\longrightarrow0.
$$
This quantitative form is what allows a single large-measure <set> to work for every tolerance, rather than choosing a different exceptional <set> for each point.
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