Darboux integral

Def 27.3

for partition $\delta \subset [a,b]$

1. ${\Delta }_j=x_{j-1}-x_j$
2. ${\inf }_{I}f={\inf }_{x\in I}f(x)$ and ${\sup }_{I}f={\sup }_{x\in I}f(x)$
3. ${\underline{\sum }}_{\sigma }={\sum }_{j=0}^{n-1}{\Delta }_j{\inf }_{I_j}f$ and ${\overline{\sum }}_{\sigma }={\sum }_{j=0}^{n-1}{\Delta }_j{\sup }_{I_j}f$ where $I_j=[x_j,x_{j+i}]$

Thm 29.1

Then we can take $I={\lim }_k{\overline{\sum }}_{{\tau }_k}f$ and ${\tau }_k=\{x_j^k\mid 0\le j\le 2^k\}$ where ${\tau }_k\subset {\tau }_{k+1}$.
Indeed we can say ${\overline{\sum }}_{{\tau }_{k+1}}\le {\overline{\sum }}_{{\tau }_k}$ and $({\overline{\sum }}_{{\tau }_k})$ is bounded below and monotone decreasing, which imply it is convergent.
Remarks:

1. $x_j^k$ means $j$-th element in $x^k$ partition
2. $x^k$ partition means we evenly split it into $2^k$ elements

Lemma 28.2

For two partition $\sigma \subset \tau$

$${\underline{\sum }}_{\sigma }f\le {\underline{\sum }}_{\tau }f\le R_{\tau }(R,\xi )\le {\overline{\sum }}_{\tau }\le {\overline{\sum }}_{\sigma }f$$

where $R$ is Riemann integral
Basically, if the largest sub-interval is larger, we will get more inaccurate result.

Lemma 28.3

For two partition $\sigma \subset \tau$ , $M=\sup (f(x)-f(y))$ which is the worst difference where $a\le x,y\le b$, we can say ${\overline{\sum }}_{\tau }\le {\overline{\sum }}_{\sigma }f\le {\overline{\sum }}_{\tau }+|\tau |M(\#\tau -\#\sigma )$ where $\#$ is the number of that set.