How can I show that the numeric sequence of a partition is equivalent to the definition of the Riemann sum?
If we assume that $f: [a,b] \to \mathbb{R}$ is bounded, how can I show these two are equivalent?
- For any $\epsilon > 0$, there is partition $P$ of $[a,b]$ such that $U(P,f) - L(P,f) < \epsilon$.
- There exists a sequence of the partition $(P_n)$ of $[a,b]$ where $\lim_{n\to\infty}(U(P_n,f)-L(P_n,f))=0$.
I think that this really isn't about the partition itself but the focus is more on how the sequences relate to the definition. My idea is to go from 1 to 2, assume 1 and say that if P were to be a sequence of $\frac{1}{n}$ length each that taking the limit of each gets the same limit making their difference 0. But I do not know how I can assume that using just 1.
For the other direction, if we assume 2, is this easier in that we can simply write out the definition of 1 and claim it to be true?
Proof:
$1) \Rightarrow 2)$:
We denote $\epsilon_{n}:=\frac{1}{n}$ with $n \in \mathbb{N}$.
Then we have that exists $\{P_{n}\}_{n \in \mathbb{N}}$ a sequence of partitions that verifique $U(P_{n},f)-L(P_{n},f)<\frac{1}{n},\forall n \in \mathbb{N}$.
In particular we have that $\lim_{n \rightarrow \infty}(U(P_{n},f)-L(P_{n},f)) \leq \lim_{n \rightarrow \infty} \frac{1}{n}=0$.
And how $U(P,f)-L(P,f) \geq 0$ we have that $\lim_{n \rightarrow \infty}(U(P_{n},f)-L(P_{n},f))=0$.
$2) \Rightarrow 1)$:
How $\lim_{n \rightarrow \infty}(U(P_{n},f)-L(P_{n},f))=0$ we have that $\forall \epsilon>0$ exists $N_{\epsilon} \in \mathbb{N} : \text{if } M > N_{\epsilon}$ then $U(P_{M},f)-L(P_{M},f))< \epsilon $.
So if we take a partition $P_{M}$ with $M > N_{\epsilon}$ we have that $U(P_{M},f)-L(P_{M},f)< \epsilon$.
Hints:
"$\implies$": We have in particular for each $n\in\mathbb N$ a partition $P_n$ such that $U(P_n,f)-L(P_n,f)<1/n$.
"$\impliedby$": Let $\epsilon>0$. Since $\lim_{n\to\infty}(U(P_n,f)-L(P_n,f))=0$, you can use the definition of a limit to find a partition that satisfies $1$.