# Seemingly faulty proof of the Integrability Criterion

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!!

• Your $n$ and $||P||$ are linked together via $n||P||\approx b-a$ and thus you can't use $n$ in expression for $\delta$. – Paramanand Singh May 16 at 10:13

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 $$M$$,
$$\|P\| < \delta = \frac{\epsilon}{2nM}- \frac{\epsilon}{2n^2M} = \frac{\epsilon}{2M}\frac{n-1}{n^2}$$
$$b-a = \sum_{j=1}^n (x_j - x_{j-1})\leqslant n\|P\|< \frac{\epsilon}{2M}\frac{n-1}{n}< \frac{\epsilon}{2M}$$
This implies that $$\epsilon > 2M(b-a)$$ contradicting the assumption that $$\epsilon$$ is arbitrary.