Proving a function is integrable over Jordan Region

[SNOOTCHIEBOOCHEE]

Homework Statement

Let E = [0, 1] × [0, 1] be a Jordan region and f : E --> R be defined as
f(x, y) = x + y.

By using the definition 12.17, show that f is integrable on [0, 1]×[0, 1]. I.e., form proper
grids to prove the integrability.

Homework Equations

Definition 12.17 (i shortened it):

F is integrable on E if for all epsilon >0 there exists a grid G such that

U(f,g)- L(f,g)< epsilon

(U(f,g) is the upper sum)

The Attempt at a Solution

Ok i honestly don't know how to define a grid G.

This seems like a really simple question unless there is something i am not getting, i figure the best way to approach this problem is just to come straight out and define a grid g.

Edit: wait. i think that the sup will always be 1 over this entire region, which means that U(f,g) <1 ... but i don't know where to go from there,

 

[SNOOTCHIEBOOCHEE]

can anybody help me?

 

[Dick]

You want to split the domain into rectangles in such a way that the upper sum minus the lower sum can be made arbitrarily small if you use enough rectangles. Split [0,1] into n parts so you are splitting the region into n^2 squares. The problem is simple enough that there are two ways to go about this. You could actually find explicit formulas for the upper and lower sum and show they have a common limit. I don't recommend this. It involves details that aren't important. The other way is to show that the difference between the maximum and the minimum of the function on each square is bounded by a constant times (1/n). Suggest a constant. The area of each (1/n)x(1/n) square is 1/n^2. That would mean that the difference between the upper and lower contribution to the sum on each square is bounded by your constant times 1/n^3. Now there are n^2 squares. Can you show the difference between the upper and lower sum goes to zero as n goes to infinity?