# Equivalence of two statements about Upper and Lower Sums

I need some help with an exercise for my analysis course. Consider $Z=\{x_0,x_1,...,x_n\}$ as a Partition of the interval $[a,b]$ and upper and lower sum of a function $f: [a,b] \rightarrow ℝ$ are defined as $$O(f,Z):=\sum_{i=1}^n sup[f(\zeta)](x_i-x_{i-1})$$ $$U(f,Z):=\sum_{i=1}^n inf[f(\zeta)](x_i-x_{i-1})$$

Show that the following statements are equivalent:

a) $\forall\varepsilon>0 \exists Z: O(f,Z)-U(f,Z)<\varepsilon$

b) $\forall\varepsilon>0 \exists\delta>0\forall Z: [|Z|<\delta \implies O(f,Z)-U(f,Z)<\varepsilon]$

The direction $b) \implies a)$ is trivial in my opinion. For the other direction it seems I have not enough information to prove it.

What you need is this geometric lemma.

Lemma. Given $\epsilon>0$ and a partition $\mathbf{P}$ of $[a,b]$, there is a $\delta>0$ such that if $\mathbf{Z}$ is another partition with $|\mathbf{Z}|<\delta$, then the total length of the subintervals of $\mathbf{Z}$ that are not contained in a subinterval of $\mathbf{P}$ is less than $\epsilon$.

Proof. If $N$ is the number of points in $\mathbf{P}$, take $\delta=\epsilon/N$ $\qquad\square$.

Now, when comparing the sums with respect to $\mathbf{P}$ and $\mathbf{Z}$, there are two types of subintervals of $\mathbf{Z}$:

1. Those containing points of $\mathbf{P}$, but whose total length is less that $\epsilon$.
2. Those contained in subintervals of $\mathbf{P}$, what allows one to make estimates on $U(f,\mathbf{Z})-O(f,\mathbf{Z})$.

Split $U(f,\mathbf{Z})-O(f,\mathbf{Z})$ into two sums. Let $M$ be a bound of $f$. On subintervals of type 1. the sum is bounded by $2\,M\,\epsilon$. Let $I$ be a subiberval of $\mathbf{Z}$ of type 2. and let $J$ be a subinterval of $\mathbf{P}$ containing $I$. Then $$0\le\sup_{I}f-\inf_{I}f\le\sup_{J}f-\inf_{J}f.$$ This should be enough to allow you to finish the proof.

• Okay, I understand the Lemma now. I've tried to split up the sums ech in the parts where I sum over all intervals which are not contained in subintervals in $P$ and over those who are contained in subintervals of $P$. I also know the the function is bounded. – hardy Apr 3 '17 at 19:38
• I have added some more hints to finish the proof. – Julián Aguirre Apr 4 '17 at 10:23