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I'd like someone to check my proof. The outline is that I'm trying to construct a partition $P$ of $[a,b]$ such that $U(P)-L(P)\le \epsilon$.

Let $\epsilon ' = \frac{\epsilon}{b-a}$

First, we construct a partition $P'$ in the following manner:

  • Pick $a$ as the first element of $P'$.
  • Compute $\delta$ such that for all $x\in[a,b]$ where $|a-x|<\delta$ it is true that $|f(x)-f(a)|<\epsilon'$. (We can do that because $f$ is continuous)
  • Now, the next element of the partition $P'$ is $a+\delta$.
  • Repeat the above procedure to compute the rest of the partition $P'$.

Now, it must be that $$U(P')-L(P') \le \epsilon' \delta_{max} N$$ where $N$ is the number of points in $P'$.

We can see that, $$N \le \frac{(b-a)}{\delta_{min}}.$$ So,

$$U(P')-L(P') \le \frac{\epsilon' \delta_{max} (b-a)}{\delta_{min}}.$$

Or

$$U(P')-L(P') \le \frac{\epsilon \delta_{max}}{\delta_{min}}.$$

So now we construct $P$ by breaking up $[a,b]$ into intervals of size $\delta_{min}$.

Is this correct and if not, where am I mistaken? One concern I have is that maybe the partition $P'$ can't be constructed in this manner without ending up with an infinite partition.

2 Answers 2

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Indeed, for all $c \in [a,b]$ you can find $\delta_c > 0$ such that for $\vert x - c \vert < \delta_c$ you have $\vert f(x) - f(c) \vert < \epsilon^\prime$.

Because $[a,b]$ is compact you can extract from the set of open intervals $$\{(c-\delta_c,c+\delta_c) \ ; \ c \in [a,b]\}$$ a finite subset that covers $[a,b]$. Then with this finite subset you can build as you did a partition $P^\prime$.

The compacity of $[a,b]$ is essential in the argument. As a counterexample, consider the continuous function $x \mapsto \frac{1}{1-x}$ defined on $[0,1)$. You wont be able to find a partition $P^\prime$ with the required properties.

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Hint: You need to use the fact that a continuous function on a compact set is uniformly continuous. Hence there is a $\delta$ that works for all the gaps.