Iterated Integrals II
Let’s get to proving the assertions we made last time, starting with
where is a bounded function defined on the rectangle .
We can start by defining
And we easily see that , where is the supremum of on the rectangle , so this is a bounded function as well. Thus the upper integral
and the lower integral
are both well-defined.
Now if is a partition of , and is a partition of , then is a partition of into subrectangles . We will define
Clearly, we have
and so we find
We also define and to be the infimum and supremum of over the rectangle , which gives us the inequalities
and from here we find
Summing on both and , and sing the above inequalities, we get
and since this holds for all partitions , the assertion that we’re trying to prove follows.
The second assertion from last time can be proven similarly, just replacing by the lower integral over . And then the third and fourth assertions are just the same, but interchanging the roles of and . Finally, the last assertion is a consequence of the first four. Indeed, if the integral over exists, then the upper and lower integrals are equal, which collapses all of the inequalities into equalities.