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I may have discovered a quick way of proving the following Theorem, but the proof is quite short and hence `too good to be true.' Could you please take a look at it and see if it makes sense?

THEOREM Suppose that the function $f:[a,b]\rightarrow R $ is bounded. Then $f:[a,b]\rightarrow R$ is integrable if and only if for each postivie number $\epsilon$ there is a positive number $\delta$ such that

$U(f,P)-L(f,P)<\epsilon$ whenever $P$ is a partition of $[a,b]$ such that $||P||<\delta$.

Before I explain my proof, some notes on the notations:

$U(f,P)$ and $L(f,P)$ denotes the upper Darboux sum and the lower Darboux sum of the function $f$ based on the partition $P$ of $[a,b]$, respectively.

$||P||$ denotes the ``gap'' of the partition $P$, that is, the length of the longest subinterval induced by $P$.

$M_i={sup}\{f(x)|x \in [x_{i-1},x_i]\}, m_i={inf}\{f(x)|x \in [x_{i-1},x_i]\}$

Now here's my proof. The (if) part is quite trivial, so I did not include it.

PROOF (only if) Let $\epsilon>0$. Suppose the function $f:[a,b]\rightarrow R$ is integrable. Since $f:[a,b] \rightarrow R$ is bounded, there exists a positive number $M$ such that $-M\leq f(x)\leq M$ for every $x$ in $[a,b]$. Let $P=\{x_0, ..., x_n\}$ be a partition of $[a,b]$. Then we have the following estimate

$ U(f,P) - L(f,P) \leq 2nM \times ||P||. $

Once the estimate is proven, we can choose $\delta=\frac{\epsilon}{2nM} - \frac{\epsilon}{2n^2M}$. Then if $P$ is a partition of $[a,b]$ such that $||P||<\delta$, we have $U(f,P)-L(f,P) \leq \epsilon-\frac{\epsilon}{n}<\epsilon $. It follows that the function is integrable.

It remains to verify the estimate. $$U(f,P)-L(f,P)=\sum_{i=1}^{n}{(M_i-m_i)(x_i - x_{i-1})} \leq \sum_{i=1}^{n}{(2M)||P||}=2nM \times ||P||. $$ Hence the estimate is proven. QED.

Thanks a lot for reading this!!

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  • $\begingroup$ Your $n$ and $||P||$ are linked together via $n||P||\approx b-a$ and thus you can't use $n$ in expression for $\delta$. $\endgroup$ – Paramanand Singh May 16 at 10:13
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You are assuming that for any $\epsilon > 0$ no matter how small, you can select a partition $P = (x_0, x_1, \ldots, x_n)$ of $[a,b]$ such that for fixed $M$,

$$\|P\| < \delta = \frac{\epsilon}{2nM}- \frac{\epsilon}{2n^2M} = \frac{\epsilon}{2M}\frac{n-1}{n^2}$$

However, this implies that

$$b-a = \sum_{j=1}^n (x_j - x_{j-1})\leqslant n\|P\|< \frac{\epsilon}{2M}\frac{n-1}{n}< \frac{\epsilon}{2M}$$

This implies that $\epsilon > 2M(b-a)$ contradicting the assumption that $\epsilon $ is arbitrary.

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  • $\begingroup$ Oh snap! I knew it was too good to be true.. thanks a lot. I really appreciate it :) $\endgroup$ – Ko Byeongmin May 16 at 5:48
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    $\begingroup$ @KoByeongmin: You're welcome. I admire your independent thinking here. $\endgroup$ – RRL May 16 at 5:50

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