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#2
Given
i think when the function f is monotonic increasing,that can be done, but otherwise?
I am the greatest magician this century!!!
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3 years late, but might be useful to people looking up the same thing.
First, note that your second partition is a refinement of the first with one extra point -- namely, y. I prove the general case for any two partitions P and P* where P* is a refinement of P, so that your result follows as an immediate corollary.
Let P be a partition of [a,b], with U(f,P) and L(f,P) being the upper and lower Darboux sums of f with respect to the partition P. Write the original partition P as
and the partition P* as
where x* is the extra point. Then:
But note that
and
substituting, we see that the RHS of the inequality is greater than or equal to 0, thus
.The inequality
is done similarly. Thus, by repeated application of this proof, this holds true for any refinement of P. Using this result, let
be a common refinement of P and Q. Then:
and
but, obviously
hence
QED.
Last edited by zetafunc (2014-05-25 19:12:41)
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