What is the proof of riemann integral?

[mathwonk]

my apologies haelfix, I've been acting like an A**H***.

Some of the things i said may nonetheless be correct however. (in addition to this.)

 

[Haelfix]

I think we've all been a tad hasty and misunderstanding one another here, again my apologies. I was referring to Rudin (my book) when I said 'rather than use sources others don't necessarily have' and pulled out the weblink, I personally have no problems whatsoever and welcome sources outside the internet.

I really don't know much about the so called Gauge integral, I appreciate your comments, they are much more informed than what I know of the subject. I also think we got a little hasty on the terminology and started talking by one another. Physicists often use the word 'pole' to mean a place where something goes wrong, not necessarily the complex analysis point of view, again my fault since I am in the wrong forum. Other than that, I agree with everything in your last posts, all of that makes perfect sense to me.

I'd like to point out that IMO from a historical point of view, I think the Lebesgue integral's primary draw is not just that it can make sense of integration outside of the reals, but also that it has various nice convergence properties that the Riemann integral does not have. Rudin talks about some of them.

 

[mathwonk]

thank you. i agree with your comment on convergence properties. that is certainly to me a key virtue of lebesgues method.

i am not sure what use you wish to make of new integration theories but there may be some out there that would interest you.

i am myself not up on these but there is a very new one (to me) in algebraic geometry called "motivic integration", and of course we are still challenged to make sense of feynman integrals.

there is a lot of work on this area, but i sense that it tries to get round the difficulty of making sense of them, by various means. here i am told the difficulty is that of integrating over path spaces of infinite dimensions.

 

[mathwonk]

re convergence properties, i have even found a source where the lebesgue integral is defined by this property.

i.e. a function is lebesgue integrable if and only if it is almost everywhere the pointwise limit of a sequence of (measurable) step functions, whose integrals converge to some number.

then the inetgeral is proved to be independent of thes equence of step functions and to depend only on the a.e. pointwise limit, and we're off.

have a good day!

 

[Hurkyl]

Hrm, that definition sounds like it would apply to (sin x)/x over [0, ∞)... just have the n-th approximation be zero on [n, ∞).

 

[master_coda]

i.e. a function is lebesgue integrable if and only if it is almost everywhere the pointwise limit of a sequence of (measurable) step functions, whose integrals converge to some number.

I don't think that's correct. I've seen a definition very close to that one, but it required the integrals of the absolute values of the step functions to converge to a number. Without that stronger condition, the "if" doesn't hold.

 

[mathwonk]

thank you master coda. i was trying to simplify the statement so as not to use technical language and i vastly oversimplified it.

(I try to take it as a compliment when I am corrected in an error, since at least it means I was understood correctly!)

the correct statement seems to be that f is integrable if and only if the sequence of step maps is not only pointwise convergent to f almost everywhere, but also "Cauchy in the L^1 norm", which as you know, means that the integrals of the absolute values of the differences |fn-fm| of the step functions, converge to zero as both n,m go to infinity.

This mdefinition makes sense even for Banach space valued functions, and may be called the bochner (version of lebesgues) integral.

It is not I think appropriate only to assume the integrals of the absolute values of the step functions converge, as that would not imply the step functions are L^1 Cauchy.

for instance we would like to be able to compute the integral of the limit from the limit of the integrals of the step functions. but suppose the nth step function were equal to 1 on the interval [n,n+1] and zero elsewhere. then all the integrals of all the absolute values would equal the integrals of the functions themselves and would all be 1. the constant sequence 1,1,1,1... certainly converges, but the pointwise limit function of these step functioins is the zero function, whose integral is not 1, i.e. is not the limit of the integrals of the approximating step functions.

so one needs also the step functions to be Cauchy in the L^1 sense, it seems.

what do you think of this, hurkyl?

the point of course is to make the lebesgue integrable functions the completion of the step functions in the L^1 norm. thus the convergence properties are "built in".

now what about cos(x)/x?

 

[master_coda]

the correct statement seems to be that f is integrable if and only if the sequence of step maps is not only pointwise convergent to f almost everywhere, but also "Cauchy in the L^1 norm", which as you know, means that the integrals of the absolute values of the differences |fn-fm| of the step functions, converge to zero as both n,m go to infinity.

I'm not sure if that's strictly necessary, although you're probably right. I thought that you only needed stronger convergence if you actually wanted the integral of the limit to equal the limit of the integrals (which is obviously much more useful, but just not what I was thinking about).

 

[mathwonk]

master coda, are you perhaps thinking of convergence of series of functions? as opposed to sequences?

 

[master_coda]

master coda, are you perhaps thinking of convergence of series of functions? as opposed to sequences?

I was thinking of sequences, but series does make a lot more sense. The combination of your definition and the vague memory of another definition seems to be confusing me.