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Integrable function

by reza
Tags: function, integrable
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reza
#1
May5-07, 08:32 AM
P: 26
how can we prove that if F(function) is integrable [a,b] then f must be bounded on [a,b]
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quasar987
#2
May5-07, 10:06 AM
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By definition, a function in integrable if the lower integral equals the upper integral. What happens to the upper integral if a function is not bounded, say, above?
Crosson
#3
May5-07, 12:53 PM
P: 1,295
By definition, a function in integrable if the lower integral equals the upper integral. What happens to the upper integral if a function is not bounded, say, above?
This, the Darboux definition, is equivalent to the Reimann-Stieltjes definition, which the OP may prefer to work with for "class" reasons.

The easiest way to see this result is by the contrapositive. If f is unbounded, the integral of f does not exist.

mathman
#4
May5-07, 07:31 PM
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Integrable function

Something is missing here. x-1/2 is integrable between 0 and 1, but it is not bounded.
Crosson
#5
May6-07, 01:27 AM
P: 1,295
Integration is defined only for closed intervals, an improper integral is an extension of this.
reza
#6
May7-07, 11:59 PM
P: 26
can you give me a mthemathica proof
mathman
#7
May8-07, 04:17 PM
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Integration is defined only for closed intervals, an improper integral is an extension of this.
This would be true if integration is defined by Darboux or ordinary Riemann. There are other approaches, Lebesgue or generalized Riemann, where boundedness would not be required.
mathwonk
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May9-07, 10:07 AM
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there are various definitions of the integral, that apply to different classes of funtions.

riemann's definition applies only to bounded ones, i.e. the limit of riemann sums is finite and independent of choice of partitions and choice points, only if the function is bounded.

this is easy to prove as follows: if f is unbounded on [a,b], then for any partition it is possible to choose a choice point so that the product of the value there by deltax will be as large as desired.

e.g. in your example of 1/x^(1/2), if you subdivide [0,1] say by intervals of length 1/n, then in the interval [0,1/n], select your point to be 1/n^4. then f(x)deltax will be equal to n. then the riemann sums have no limit.


what you are thinking of in this case is called the improper integral, a different definition that applies when the function is only unbounded locally near finite number of points. then here e.g. the improper integral ius defined differently, as the limit of the riemann integrals on intervals where the function is bounded, say [e,1], as e goes to 0.

riemann himself mentioned this extension in his paper where he defined the usual riemann integral.

other more flexible definitions by lebesgue and others apply to even more functions, such as the function which equals 1 on the rationals and 0 on the irrationals.

in this case one can adapt the idea of improper integrals as follows. approximate the function rather than the interval, by the sequence of functions fn where fn equals 1 on those rationals whose denominator is no larger than n, and 0 elsewhere.

then one gets that the limit of these integrals is zero, a reasonable definition for the integral of the original function. the technical nuisance here is proving independence of the choice if the approximating sequence of integrable functions.

lebesgues own approach, although equivalent is more complicated, and istead assigns a length separately to the sets of rationals and irrationals. since in his theory the set of rationals has length zero, the integral is again zero.


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