Doubt in Proof of "Measurable Function is Pointwise Limit of Simple Functions"

Question

The proof for reference, I think that one can understand the proof similar to the answer here but taking $f_n(x)$ as binary n-digit approximation to x. Now, for each positive integer $n$ and for $k$ = $1\,, 2\,, \ldots, n^2$, if we let $A_{n,k}$ = $\{x \in A: \frac{k-1}{n} \leq f(x) \lt \frac{k}{n}\}$ and, defining $f_n(x) = \frac{k-1}{n}$ at each point in $A_{n,k}$ and $n$ at points in $A - \cup_{k} A_{n,k}$. Would the argument after appropriate changes work for pointwise convergence?

Answer 1

I believe if you take any appropriately small mesh such as $\Delta_n = 1/n,1/2^n,1/n^\epsilon$, e.g., and define the approximation $$ f_n(x) = \sum_{k\in\mathbb Z} f(k\cdot\Delta_n)\chi_{[k\cdot\Delta_n,(k+1)\Delta_n)}(x), $$ then the resulting step functions will converge to $f(x)$ pointwise almost everywhere. The main reason that the dyadic mesh $\Delta_n = 2^{-n}$ is chosen is so that we can verify the sequence of approximants is increasing in $n$. This becomes simple to verify for the dyadic mesh because of the neat combinatorics of nested dyadic intervals.

It's desirable to know there is an increasing sequence of approximants because in applications we want to be able to verify propositions about general functions using step functions, and then extend those propositions to general integrable functions using the monotone convergence theorem.

You can read an example attempt to change the mesh and the problem for the resulting sequence of approximants to be increasing on this stackexchange page.