Proof: Every Positive Measureable Function There Exist Simple Measureable Function

**BradM707** · December 30th 2010, 01:10 PM

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

Proof:
For $n=1,2,3\ldots$ and for $1\leq i \leq n2^n$ , define
$E_{n,i}=f^{-1}([\frac{i-1}{2^n},\frac{i}{2^n}))$ and $F_n=f^{-1}([n,\infty])$ and put $s_n=\sum_{i=1}^{n2^n}\frac{i-1}{2^n}\chi_{E_{n,i}}+n\chi_{F_n}$ where $\chi_{E_{n,i}},\chi_{F_n}$ 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 $s_1 \leq s_2 \leq \ldots$ 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 $f(x)=\infty$ then $s_n(x)=n$ and we're good. If $f(x)<\infty$ then we should be able to show that $s_n(x)\geq f(x)-2^{-n}$ for large enough n. I am having problems seeing how to actually arrive at this result.

Any help would be appreciated,
Brad

**Bruno J.** · December 30th 2010, 04:23 PM

Fix $n$ . Take any $x \in X$ Then $x$ belongs to exactly one of $E_{n,i}$ or $F_n$ . If $x \in E_{n,i}$ , for example, this means that $\frac{i-1}{2^n}\leq f(x) < \frac{i}{2^n}$ . But $s_n(x) = \frac{i-1}{2^n}$ . Hence $s_n(x) \leq f(x)$ .

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