riemann integrability
lets see what we have here. suppose we can cover the set of discon tinuities by a sequence of intervals whose total length is less than e. then some finite number of those intervals covers most of the discon tinutities and has total length e./2,.. is our funbtion bounded? i guess so, so we have a bounded function and given e we can
hmmmm this requires a little thought. maybe i need to cheat and actually use an idea, due to riemann, which fortunately i remember, called oscillation,
the oscillation of f at a point c is the limit of the lengths of the smallest intervals that contain the images of the numbers f(c-a,c+a) as a goes to zero. i.e. let (c-a,c+a) be an interval shrinking down to the point c, as goes to zero. the set of vaklues f takes on this interval lies in some smallest bounded interval.
as a goes to zero these "image intervals" are bounded, but shrink down to zero length if and only if f is continuous at c.
so riemann introduced this measure of how discontinuous f is at c. now the idea is to consider the set of points c where the oscillation is more than e, where e is a given positive number. then using compactness one shows that this set has "content zero", i.e. can be covered by a finite set of intervals, of total length less than e. then one proceeds like this.
i.e. suppose we want upper and lower riemann sums of f on an interval [a,b], that are closer together than e. If f is bounded by -M and M, choose a finite set of intervals that cover all the points of [a,b] where f has oscillation more than e/6M, and of total length less than e/6M.
Then on these intervals f may have ocillation as much as 2M, but the total lengths of those intervals is less than e/6M, so these parts of the graph of f can be covered by rectangles of total area les than let's see...
uh...e/3?
then off these intervals at all the points f has oscillation less than e/6M, so there exists rectangles covering those points with upper and lower sums differeing by less than say e/3M. so this gives upper and lower sums with total difference les than e.
i am going to stop now, because i have done a little example.
what i remember is this:
this proof appears in spivaks calculus on manifolds, and in riemanns works in his paper on trigomometric series.
i can say the last time i recall doing this proof was 1970, driving across country and bored, so i worked it out in my head in the car.
if this seems impressive as a feat of memory, remember this is what we do all the time.
as charles barkley said an average player has no chance against a pro because they play every day.
or let me give an example closer to home. my son was an all state basketball player in high school and led the nation for a while in college in 3 point shooting at 59 %.
I thought this was great. But he got no chances at all to play pro in the US but was a part time semi pro in Europe. But when he came home at xmas, while he was playing as a semi pro, I watched him play, and he was much better than he ever was in high school or college.
He almost never missed. Warming up, i saw him hit about 15-18 straight mid range jumpers.
And he said he had been at a camp with an NBA player from utah, who was suppose to hit 10/12, 3 - pointers or everyone had to run laps.
the pro missed the first two then hit 10 in a row. jeff hornacek, that was it.
people who play every day are so much more accurate than we are there is no comparison.
math professors are profesional mathematicians. still some of us are not as sharp as others.