 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. HenceThese are inclusions, not necessarily strict inclusions. The displayed sets describe eventual uniform error tolerances at an individual point.

Fix $k\ge 1$ and $x\in A$. By pointwise convergence, there is an $N=N(x,k)$ such that $|f_m(x)-f(x)|\le 1/k$ for every $m\ge N$. This is exactly $x\in E_N^{(k)}$, provingTogether with monotonicity in $n$ and finite measure of $A$, continuity from below of a measure gives $\lambda (E_n^{(k)})\uparrow \lambda (A)$. ConsequentlyThis 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.