Approximation of $f\in L_p$ with simple function $f_n\in L_p$ Let us use the definition of Lebesgue integral on $X,\mu(X)<\infty$ as the limit$$\int_X fd\mu:=\lim_{n\to\infty}\int_Xf_nd\mu=\lim_{n\to\infty}\sum_{k=1}^\infty y_{n,k}\mu(A_{n,k})$$where $\{f_n\}$ is a sequence of simple, i.e. taking countably (infinitely or finitely) many values $y_{n,k}$ for $k=1,2,\ldots$, functions $f_n:X\to\mathbb{C}$ uniformly converging to $f$, and $\{y_{n,k}\}=f_n(A_{n,k})$ where $\forall i\ne j\quad A_{n,i}\cap A_{n,j}=\emptyset$.
I know that if any sequence $\{f_n\}\subset L_p(X,\mu)$, $p\geq 1$ uniformly converges to $f$ then it also converges with respect to the norm $\|\cdot\|_p$ to the same limit, which is therefore an element of $L_p(X,\mu)$.
I read in Kolmogorov-Fomin's Elements of the Theory of Functions and Functional Analysis (1963 Graylock edition, p.85) that hence an arbitrary function $f\in L_2$ can be approximated [with respect to norm $\|\cdot\|_2$, I suspect] with arbitrary simple functions belonging to $L_2$.
I do not understand how the last statement is deduced. If it were true, I would find the general case for $L_p$, $p\geq 1$, even more interesting. I understand that if $f\in L_p\subset L_1$ then for all $\varepsilon>0$ there exists a simple function $f_n\in L_1$ such that $\forall x\in A\quad |f(x)-f_n(x)|<\varepsilon$ and then, for all $p\geq 1$, $\|f-f_n\|_p<\varepsilon$, but how to find $f_n\in L_p$ (if $p=2$ or in general $p>1$)?
Can anybody explain such a statement? Thank you so much!!!

In many sources, simple functions are those measurable functions that have a finite set of values. But here a countably infinite set of values is allowed.  This allows one easily approximate any measurable function $f$ uniformly by simple functions: for example, let
$$
f_n(x) = \frac{\lfloor n f(x)\rfloor }{n}
$$
and observe that $f-\frac{1}{n}\le f_n\le f$ everywhere. On a finite measure space, this implies $f_n\to f$ in $L^p$ norm for every $p\in [1,\infty]$.
