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Theorem 2.2.5 in Friedman's Modern Analysis Book:

Let $f$ be a non-neg measurable function. Then there exists a mono-increasing sequence $\{f_n\}$ of simple non negative functions such that $\{f_n\}$ converges to $f$ everywhere.

Proof: For each integer $n\geq 1$ and for each $x\in X$, let \begin{align} f_n(x)=\left\{ \frac{i-1}{2^n}, \quad\quad\quad\quad \text{ if } \frac{i-1}{2^n}\leq f(x)\leq \frac{i}{2^n}, i=1,2,...,n2^n,\\ n, \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{ if } f(x)\geq n \right. \end{align} Then the $f_n$ are simple non-neg functions, and $f_{n+1}\geq f_n(x)$. At a point $x$ where $f(x)<\infty$, \begin{equation} 0\leq f(x)-f_n(x)\leq \frac{1}{2^n} \quad \quad \text{ if } n> f(x) \end{equation} At a point $x$ where $f(x)=\infty, f_n(x)=n$. This shows that, for any point $x, f_n(x)\rightarrow f(x)$ as $n\rightarrow\infty$

Ok. I am struggling with seeing how this function is a simple function, and how one would know it come up with such a function. I see sequences like $\frac{1}{2^n}$ used a lot in integration theory, probably because of how they converge nicely, but I never know when to use things like this. If we are summing from over $n$ then it makes sense to use it, since it is a series that converges nicely, but I don't see the use here. I also don't see exactly why this even defines a simple function.

Thanks for the help!

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For every $n$, $f_n$ only takes a finite amount of values, $n2^n+1$ to be exact. They do depend on $f$, yes, but $f$ is fixed at that point. Since the value of $f_n(x)$ doesn't involve $x$, it is constant. Hence it is a simple function.

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  • $\begingroup$ Ok I see your point... but what about the how the definition of a simple function talks about the value of a point equals some constant if $x\in E_i$ and equals zero if $x \notin \bigcup E_i?$ Is it even possible for $f$ to take a value of zero here? And what the the $E_i$'s here? Thanks! $\endgroup$ Jan 4, 2017 at 8:36

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