# Is it necessarily the case that $\lim_{k\to\infty} \text{gap } P_{k} = 0$ if $\{P_{k}\}$ is an Archimidean sequence of partitions?

Is it necessarily the case that $$\lim_{k\to\infty} \text{gap } P_{k} = 0$$ if $$\{P_{k}\}$$ is an Archimidean sequence of partitions?

I know that by the definition of an Archmedean sequence of partitions, we have

$$\lim_{k\to\infty}[U(f, P_{k}) - L(f, P_{k})] = 0.$$

Also by Archimedean Riemann, we get

$$\lim_{k\to\infty} L(f, P_{k}) = \int_{I} f = \lim_{k\to\infty} U(f,P_{k}).$$

So it's nowhere stated, but I can't come up with a counterexample. Any help is appreciated.

• What if $f$ is constant on some non-degenerate interval? – David Mitra Apr 23 at 10:37
• If $f$ is constant on a non-degenerate interval, we get $U(f, P_{k}) = c \cdot \text{vol } J$ and $L(f, P_{k}) = c \cdot \text{vol } J$. I see that the limit goes to $0$ then, but do I need to provide an example of a sequence $P_{k}$? – user666729 Apr 23 at 23:31
• Maybe I'm misinterpreting "gap"; I took it to mean the length of the largest subinterval of the partition. (Then take each $P_k$ to be the entire interval.) – David Mitra Apr 24 at 2:43