0

I do not know how to calculate using the definition of upper and lower sum the following sums of the function to segments:

Supose $a<c<b$, $f(c)=1$ and $f(x)=0$ for all $x \in [a,b]-\{c\}$. Compute the lower and upper integrals of $f$.

It confuses me not to know if there is any relationship between a, b and c that influences the calculation or how to correctly use the notion of Darboux's sum. Any help would be appreciated.

Hendrik Matamoros
  • 659
  • 1
    You can read some here – Jakobian Jul 12 '18 at 18:12
  • I could infer that the lower integral was gonna be equal to zero no matter the partition and the upper integral was always gonna be greater or equal to zero. But I would not know how to express these calculations formally using the definition of upper and lower sum respectively. – Hendrik Matamoros Jul 12 '18 at 18:12

1 Answers1

1

Let $a=x_0<x_1<\dots<x_{n-1}<x_n=b$ be a decomposition $P$ of $[a,b]$ then then the lower Darboux sum with respect to $P$ is $$L(f,P)=\sum_{k=1}^n \inf_{x\in [x_{k-1},x_k]}f(x)\cdot (x_k-x_{k-1})=\sum_{k=1}^n 0\cdot (x_k-x_{k-1})=0.$$ On the other hand for upper Darboux sum with respect to $P$ we have two cases: $$U(f,P)=\sum_{k=1}^n \sup_{x\in [x_{k-1},x_k]}f(x)\cdot (x_k-x_{k-1})= \begin{cases} 1\cdot (x_i-x_{i-1})\quad\text{if $x_{i-1}<c<x_{i}$,}\\ 1\cdot (x_{i+1}-x_{i-1}) \text{ if $c=x_{i}$.} \end{cases}$$ Can you take it from here and find $U(f)$ and $L(f)$?

Robert Z
  • 145,942