# Proof: Every Positive Measureable Function There Exist Simple Measureable Function

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• Dec 30th 2010, 01:10 PM
BradM707
Proof: Every Positive Measureable Function There Exist Simple Measureable Function
Hello,
This is my first post on this site, so please excuse me if the post is a bit rough. I'm currently reading Rudin's Real and Complex Book and ran across Theorem 1.17

From Book:
Theorem 1.17:
Let $\displaystyle \small{f: X \rightarrow [0,\infty]}$ be measurable. There exist simple measurable functions $\displaystyle s_n$ on X such that.
(a) $\displaystyle 0 \leq s_1 \leq s_2 \leq \ldots \leq f$
(b) $\displaystyle s_n(x) \rightarrow f(x), n \rightarrow \infty, \forall x \in X$

Proof:
For $\displaystyle n=1,2,3\ldots$ and for $\displaystyle 1\leq i \leq n2^n$, define
$\displaystyle E_{n,i}=f^{-1}([\frac{i-1}{2^n},\frac{i}{2^n}))$ and $\displaystyle F_n=f^{-1}([n,\infty])$ and put $\displaystyle s_n=\sum_{i=1}^{n2^n}\frac{i-1}{2^n}\chi_{E_{n,i}}+n\chi_{F_n}$ where $\displaystyle \chi_{E_{n,i}},\chi_{F_n}$ are the characteristic functions of E and F, respectively.

Questions:
Part a) So we know that E and F are measurable sets because the pull back of a Borel set through a measurable function is a measurable set. And with a little algebra we can see that $\displaystyle s_1 \leq s_2 \leq \ldots$ How do you prove that the simple functions are less than or equal to f for all n? I can do this with finite examples, but don't know how it to extend it to an arbitrary function.

Part b) If $\displaystyle f(x)=\infty$ then $\displaystyle s_n(x)=n$ and we're good. If $\displaystyle f(x)<\infty$ then we should be able to show that $\displaystyle s_n(x)\geq f(x)-2^{-n}$ for large enough n. I am having problems seeing how to actually arrive at this result.

Any help would be appreciated,
Brad
• Dec 30th 2010, 04:23 PM
Bruno J.
Fix $\displaystyle n$. Take any $\displaystyle x \in X$ Then $\displaystyle x$ belongs to exactly one of $\displaystyle E_{n,i}$ or $\displaystyle F_n$. If $\displaystyle x \in E_{n,i}$, for example, this means that $\displaystyle \frac{i-1}{2^n}\leq f(x) < \frac{i}{2^n}$. But $\displaystyle s_n(x) = \frac{i-1}{2^n}$. Hence $\displaystyle s_n(x) \leq f(x)$.