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The original theorem is a as follows:

THEOREM Let $f$ be continuous on $[a,b]$. Then for any $\epsilon>0$ there exists a finite partition of $[a,b]$ such that the oscillation of $f$ on each subinterval is smaller than $\epsilon$.

The proof is as follows.

PROOF We argue by contradiction. Suppose there is an $\epsilon_0$ such that no finite partition of $[a,b]$ exists such that the oscillation is less than $\epsilon_0$. We use a bisection argument. Since the theorem fails on $[a,b]$, it must fail on $[a,c]$ or $[c,b]$, where $c$ is the midpoint of the interval. Call $[a_1,b_1]$ the interval where it fails. If it fails on both, we agree to choose the left interval. Continue the process, call $[a_{n+1},b_{n+1}]$ the half of the interval $[a_{n},b_{n}]$ where the theorem fails. Note that on each $[a_n,b_n]$ the oscillation must be $\geq \epsilon_0$. Let $A$ be the set consisting of $a,a_1,a_2,\dots$, and let $\alpha=\sup A$. Then $\alpha\in[a,b]$, so that $f$ is continuous at $\alpha$. Then, there is an interval $(\alpha-\delta,\alpha+\delta)$ where the oscillation of $f$ is less than $\epsilon_0$. But we must have $[a_n,b_n]\subset (\alpha-\delta,\alpha+\delta)$ for some $n$ with $(b-a)/2^n<\delta$, so that the oscillation of $f$ in $[a_n,b_n]$ is also less than $\epsilon_0$. This contradiction proves the thereom.

It has already been prove that if $f$ is continuous on $[a,b]$ then exist $c$ and $d$ on that interval such that $f(c)=\inf f $ and $f(d)=\sup f$

where $\inf f =\inf \{f(x):a\leq x\leq b\}$ and $\sup f =\sup\{f(x):a\leq x\leq b\}$. According to Apostol, we define the oscilation of $f$ on $[a,b]$ by $f(d)-f(c)$

Now, I need want to prove the part that is in boldface. Given any $\epsilon >0$ if $\alpha\in [a,b]$, then there exists a $\delta >0$ such that the oscillation of $f$ on $(\alpha-\delta,\alpha+\delta)$ can be made smaller than $\epsilon$.

He hasn't defined what the oscillation is for open intervals, and he talks about the maximum and minimum of $f$, calling them $M(f)$ and $m(f)$ to set the oscillation to be $M(f)-m(f)$. I'm guessing on closed intervals, this is $f(d)-f(c)$ as before. $(1)$ doesn't hold for open intervals, like $(\alpha-\delta,\alpha+\delta)$. How can I prove this? Any hints? What is a more concrete definition of the oscillation of $f$ on an interval $I$?

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    Yes, I thought of that.2012-09-09

1 Answers 1

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The oscillation of a function $f$ on a set $X$ is defined as:

$ \omega(f, X) = \sup_{x, y \in X} \left|f(x) - f(y)\right| $

If $f$ is continuous on $[a, b]$, then it's uniformly continuous and for each $\epsilon$ we can find a $\delta$ so that, for all $x,y$,

$ |x - y| < \delta \Rightarrow \left|f(x) - f(y)\right| < \epsilon $

In particular:

$ \forall x, y \in \left(\alpha - \frac{\delta}{2}, \alpha + \frac{\delta}{2}\right) : \left|f(x) - f(y)\right| < \epsilon $

Hence, the oscillation of $f$ on this neighborhood is also smaller than $\epsilon$.

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    @PeterTamaroff Not at all!2012-10-12