Construct a sequence of simple functions (finite linear combination of characteristic functions of measurable sets) converging to $f(x)=|\frac{1}{x}|$
I do not see how there is anything different to the this problem then constructing a sequence to a general function $f$ and just saying it holds for any $f$. There must be a caveat, however the only thing I can think to do is let $f(x)=|\frac{1}{x}|$ and reciprocate the proof for general functions. For $n=0,1,2,...$ and $0 \leq k \leq 2^{2n}-1$ let $E_n^{k}=f^{-1}(k2^{-n},(k+1)2^{-n}]$ and $F_n=f^{-1}(2^n,\infty]$. Define $\phi_n=\sum\limits_{k=0}^{2^{2n}-1}k2^{-n}\chi_{E^n_k}+2^n\chi_{F_n}$ the $\phi_n$ is an increasing sequence converging to $f$. Is there any different way to do this when given an explicit function like I have been?
Divide $y$ axis into strips $A_n$ which are $\frac{1}{n}$ wide. Let $A_{n,k}$ be the set of points where $y_{n,k-1}\le f(x)\lt y_{n,k}$. Define $f_n(x)=\frac{y_{n.k-1}+y_{n,k}}{2}$ for $x \in A_{n,k}$. As $n\to \infty$, $A_n$ intervals shrink to $0$ and $f_n(x)\to f(x)$.
Personal: this was my first exposure to Lebesgue integral.