How does one show that the fat cantor characteristic function is nonriemann int'able?


by Demon117
Tags: cantor, characteristic, function, intable, nonriemann
Demon117
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#1
Oct27-10, 08:15 PM
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I have been thinking about this for quite some time now. When I look at the function that descibes the fat cantor set namely:

f(x) = 1 for x[tex]\in[/tex]F and f(x) = 0 otherwise, where F is the fat cantor set.

I wonder, how do I prove that this is non-riemann integrable?

I have considered looking at the Riemann-Lebesgue theorem which gets me nowhere. So f is obviously bounded. But isn't this f discontinuous at all x[tex]\in[/tex][0,1]? This would imply that the discontinuity points of f need to be a zero set in order for it to be riemann integrable. But isn't the fat cantor set F not a zero set?

Any advice?
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mathwonk
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Oct28-10, 09:58 AM
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well on the next page after defining his integral, riemann proved that, in modern language, a bounded function is riemann integrable if and only if its set of discontinuities has measure zero. so look for the points of discontinuity of your function.

you might give a definition for the "fat cantor set". i presume it is called fat because it does not have measure zero.
Demon117
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Oct28-10, 01:58 PM
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Quote Quote by mathwonk View Post
well on the next page after defining his integral, riemann proved that, in modern language, a bounded function is riemann integrable if and only if its set of discontinuities has measure zero. so look for the points of discontinuity of your function.

you might give a definition for the "fat cantor set". i presume it is called fat because it does not have measure zero.
So then based on this I should just find the discontinuities of f. Then if this set of discontinuities does not form a zero set it is non-riemann integrable. Isn't that exactly what the Riemann-Lebesgue theorem states?

mathwonk
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Oct29-10, 12:04 PM
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How does one show that the fat cantor characteristic function is nonriemann int'able?


well i don't know what lebesgue had to do with it, since riemann proved it on the next page after defining his integral, in his paper on representing functions by trigonometric series. Lebesgue of course was not born for almost 10 years after riemann died. But perhaps some people with less concern for historical precedent do call this theorem as you say. I do not know.....

Well I have found a citation for this theorem crediting it entirely to lebesgue in the book by m.e. munroe on introduction to integration. perhaps he never read riemann.

Or perhaps some people do not notice that riemann's version is equivalent to this "measure zero" statement? well a similar situation holds in many historical cases such as the so called "theorem of sard" whose principal corollary was later noticed to be due earlier to a. b. brown.

So forgive me but I often fail to understand references to theorems by name. I need to know the statement.
Bacle
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#5
Nov12-10, 03:24 AM
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Why not also try to work with upper- and lower- sums, see if there values coincide.?
Bacle
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#6
Nov12-10, 04:27 AM
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The restriction to countably-many discontinuities has to see with the fact that
an uncountable sum ( of course, uncountably-many non-zero terms) cannot
converge, i.e., will necessarily be infinite.


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