Sequential Characteristic of Riemann Integral

Question

Wanted to check the validity of a proof:

Claim:

Let $f:[a,b]\to\mathbb R$ be bounded. Then f is integrable if and only if there exists a sequence of partitions $P_n$ such that

$\lim_{n\to\infty}U(f,P_n)-L(f,P_n) = 0$

My attempt at a proof:

Suppose f is integrable $\implies \forall \epsilon > 0\exists$ a partition P such that $U(f,P) - L(f,P) < \epsilon$

Let $\epsilon = \frac{1}{n} , n \in\mathbb N$

Thus for each $n \in \mathbb N$ choose $P_n$ such that

$U(f,P_n) - L(f,P_n) < \frac{1}{n} =\epsilon$

Thus $\lim_{n\to\infty}U(f,P_n)-L(f,P_n) = 0$

Then proving the contrapositive:

Suppose $f$ is not integrable

$\implies U(f) $ does not equal $L(f)$

Thus $\forall n\in \mathbb N \exists \epsilon > 0$ such that

$U(f,P_n)-L(f,P_n) > \epsilon$

Thus the claim follows.

Any help and criticism is appreciated, thanks.

Answer 1

The if part is not correct. The contrapositive would be: if $f$ is not integrable, then there is $\textit{no}$ sequence of partitions with the advertised property. It is probably easier to do it directly: you need to show that if there exists a sequence of partitions $\{P_n\}$, with the advertised property, then $\int f$ exists. For this, just note that, for each integer $n\ge 0,$

$$L(P_n,f)\le \underline \int f\le \overline \int f\le U(P_n,f)$$

Answer 2

I have come up with an idea for the implication in the other direction:

Suppose $\lim_{n\to \infty} U(f,P_n)-L(f,P_n) = 0$

$\implies \forall \epsilon > 0 \exists N \in \mathbb N$ such that

$U(f,P_n)-L(f,P_n) < \epsilon$

$\implies \forall \epsilon > 0 \exists$ a partition $Q=P_n$ such that when $n>N$

$U(f,Q)-L(f,Q) < \epsilon$

and thus f is integrable.