Assume that every $f$ is continuous on $[a,b] \times [c,d]$, since $[a,b] \times [c,d]$, $f$ is uniformly continuous, so let $u,v \in [a,b] \times [c,d]$ for all $\epsilon>0$, there exist an $\delta>0$ such that

$$|u_i-v_i|<\delta =>|f(u)-f(v)|<\epsilon$$

Let $m,n$ be positive integers, then we have partition points

$$a=x_0<x_1<x_2<...<x_m=b$$ $$c=y_0<y_1<y_2<...<y_n=d$$

let $P=\{x_1,x_2,...,x_m;y_1,y_2,...,y_n\}$ be the partition in $[a,b] \times [c,d]$. I want to show that $U(f,P)=L(f,P)$ but I don't know how to get that from here.

Note: I check a couple version on the internet, they use something about volume, and I don't think I can use that or know how to use it yet.


Denote $B=[a,b]\times [c,d]$. I take you're working with Darboux sums, so it suffices you show that for each $\varepsilon >0$ there exists a partition $P$ for which $$U(f,P)-L(f,P)<\varepsilon$$

This difference of sums will have the form $$\sum_{i=1}^n (M_i-m_i)\cdot \nu(B_i)$$

where $B_1,\ldots,B_n$ is a partition of your cell into blocks, and $$M_i=\sup_{x\in B_i} f(x)\\m_i=\inf_{x\in B_i} f(x)$$ Note we will have $M_i-m_i=\sup\limits_{x\in B_i}|f(x)-f(y)|$.

Saying $f$ is uniformly continuous means we can choose a partition of blocks $B_1,\ldots,B_n$ such that $M_i-m_i<\varepsilon {\nu}(B)^{-1}$ for each $i$: given this $\varepsilon>0$, there is $\delta>0$ such that $\lVert x-y\rVert <\delta$ gives $|f(x)-f(y)|<\varepsilon$. If we take blocks with diagonals $<\delta$, then whenever $x,y$ are in the same block, $\lVert x-y\rVert <\delta$. Then $$\sum_{i=1}^n (M_i-m_i)\cdot \nu(B_i)<\varepsilon {\nu}(B)^{-1}\sum_{i=1}^n \nu(B_i)=\varepsilon{\nu}(B)^{-1} {\nu}(B)=\varepsilon$$

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  • $\begingroup$ May I have some question please? what is $v(B_i)$? By saying block, is it the same as ball or neighborhood? $\endgroup$ – Diane Vanderwaif Apr 5 '14 at 14:00
  • $\begingroup$ @DianeVanderwaif A rectangle $B$ in the plane may be partitioned in smaller rectangles, which I call $B_1,\ldots,B_n$, "blocks". $\nu(B_i)$ denotes the area they enclose. $\endgroup$ – Pedro Tamaroff Apr 5 '14 at 14:03
  • $\begingroup$ oh, I see, $v(B_i)=\Delta x_i \Delta y_j$ for some positive integers $i,j$? One last question $\epsilon v(B)^{-1}= \frac {\epsilon}{v(B)}$? $\endgroup$ – Diane Vanderwaif Apr 5 '14 at 14:13
  • $\begingroup$ Yes ${}{}{}{}{}$ $\endgroup$ – Pedro Tamaroff Apr 5 '14 at 14:19

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