I often find it stated in Kolmogorov-Fomin's (for ex. here in the proof of ex. 2) that the sequence $\{f_n\}$ of summable simple functions uniformly converging to $f$, that are used to define the Lebesgue integral $\int_A fd\mu=\lim_n \int_A f_nd\mu=\lim_n \sum_k y_{n,k}\mu(A_{n,k})$ where $A_{n,k}=\{x\in A:f_n(x)=y_{n,k}\}$, can be chosen to be non-decreasing.
How can we verify that $\{f_n\}$ can be chosen so that it uniformly converges to $f$ while $\forall x\in A$ $f_1(x)\leq f_2(x)\leq...$ and while, for all $n$, $\sum_k |y_{n,k}|\mu(A_{n,k})<+\infty$? $+\infty$ thanks for any answer!
For constructing the integral of $f$ for $f≥0$, you don't worry if its $∞$. One explicitly writes down a formula for one choice of $f_n$, for example
$$ f_n := \sum_{k=1}^{2^{n^2}} \frac{k-1}{2^n} \mathbf{1}_{\left\{ \frac{k-1}{2^n} ≤ f < \frac{k}{2^n} \right\}} $$
Note for every $x$, only 1 summand is ever non-zero and that summand is not more than $f$. You can also convince yourself that $f_n(x)$ increases in $n$, and also by construction $f(x) - f_n(x) ≤ \frac{1}{n}$ which is a bound independent of $x$, so we have uniform convergence. Whats left to do is to show that the integral is independent of choice of approximating functions.
At this point we allow $∫f ∈ \mathbb{R}\sqcup\{∞\}$ because everything is non-negative, so tending to infinity is the only 'kind of divergence' possible. For $f$ that can take positive values as well as negative values, we only try to assign a value to $∫f$ if $∫|f|<∞ $, so the bound $∑|a_{nk}| \mu(A_{nk}) < ∞$ holds by assumption.