I am having a difficult time understanding on how to construct a sequence of non-negative simple functions $\phi_{n}$ whose limit is $f$.
A problem that I am working on is to evaluate $\int_{\Omega} f dm$ where $f=2x+1$ on $\Omega=[3,4]$. There is the following example in my textbook that I have been staring at:
Given $f:[0,1]\rightarrow\mathbb{R}$, $f(x)=x$. Evaluate $\int_{\Omega} f dm$.
They choose $\phi_{n}=\sum_{i=1}^{2^n} \frac{i-1}{2^n} \chi_{[\frac{i-1}{2^n},\frac{i}{2^n}]}$. I understand that $\phi_{n}\in S_{+}$ and $\phi_{n} \nearrow f$, but I want to know why they chose that specifically and if so, how to generalize that for my case of $f(x)=2x+1$ or $f(x)=\alpha x+\beta; \alpha,\beta\ \geq0$.
Also, is there a general method for the construction of such a sequence of simple functions not only for linear functions but for other functions?
Note: I am not looking for a solution to my posed question.
Thank you.
Evaluate $\int_\Omega f dm$ where $f=2x+1$ on $\Omega=[3,4]$.
Let $\phi_{n}=\sum_{i=1}^{2^n} \min_{[3+\frac{i-1}{2^n}, 3+\frac{i}{2^n}]}{f(x)} \cdot \chi_{[3+\frac{i-1}{2^n}, 3+\frac{i}{2^n}]}$.
Therefore,
$\int_{[0,1]} (2x+1) \cdot dm=\lim_{n\rightarrow \infty}\sum_{i=1}^{2^n} (2 \cdot (3+\frac{i-1}{2^n})+1) \cdot \frac{1}{2^n} = \lim_{n\rightarrow \infty}\frac{1}{2^{2n}} \cdot \sum_{i=1}^{2^n} (7 \cdot (2^n)+2(i-1)) = \lim_{n\rightarrow \infty}\frac{2^{2n+3}-1}{2^{2n}} = 8.$