What is the proof of riemann integral?

[semidevil]

so riemann integral pretty much says that if you take closer and closer approximations, then you can find the area of whatever(not too precise, I know, but doing the rectangles and stuff).

I'm looking for a proof of it, but all I can find are more general things, i.e, Riemann integeral is integeral on [a, b], etc etc.

how was it proven that taking approximations and all that, actually worked??

 

[HallsofIvy]

But it doesn't say that for EVERY function.

What you can do is this: inside of each interval take M to be the largest value of f(x) and m to be the smallest. for all x in that interval m< f(x)< M so m&delta;x< f(x)&delta;x< M&delta;x. That is: the "true area" inside each interval lies between the
rectangle area m&delta;x and the rectangle area M&delta;x. The "true area" under the curve lies between the sum of m&delta;x and M&delta;x.

IF the limit of sum m&delta;x and sum of M&delta;x are the SAME, THEN it is clear that the "true area" must be equal to that joint limit (the "pinching" theorem).

Of course, THAT is not always true! For example if f(x)= 1 for x irrational, f(x)= 0 for x rational, since every interval contains some rational and some irrational, the "upper sum" (sum of M&delta;x) is always 1 and the "lower sum" (sum of m&delta;x) is always 0. Those are not the same and so that function is NOT "Riemann integrable".

 

[matt grime]

It is the definition that IF the approximations using rectangles works THEN it is (Riemann) integrable. It is not something one proves.

 

[CrankFan]

so riemann integral pretty much says that if you take closer and closer approximations, then you can find the area of whatever(not too precise, I know, but doing the rectangles and stuff).

I'm looking for a proof of it, but all I can find are more general things, i.e, Riemann integeral is integeral on [a, b], etc etc.

how was it proven that taking approximations and all that, actually worked??

Tom Apostol's Calculus Volume I has pretty good coverage of what I think you're looking for. You might want to check it out of a local library and give pages 57-80 a look.

 

[semidevil]

It is the definition that IF the approximations using rectangles works THEN it is (Riemann) integrable. It is not something one proves.

so for example, this link

http://www.hyper-ad.com/tutoring/math/calculus/Construction of the Riemann Integral.html


is this pretty much it? to me, it seems like it.


and reading it, how did they conclude that the upper limit equals teh lower limit?

 

[arildno]

There is an unfortunate representation in the link, that it seems to regard the concept of "area under the curve" as well-defined, IRRESPECTIVE of whether or not the upper and lower partial sums converge to a common number in the limit.
But, it is precisely when the sums DO converge to a single, common number that it is meaningful to speak of the "area" beneath the curve, in the Riemann sense (that single, common number is then the assigned area value).
HallsofIvy's example shows a function in which an area in the Riemann sense of the word is impossible to assign to the curve.

 

[semidevil]

There is an unfortunate representation in the link, that it seems to regard the concept of "area under the curve" as well-defined, IRRESPECTIVE of whether or not the upper and lower partial sums converge to a common number in the limit.
But, it is precisely when the sums DO converge to a single, common number that it is meaningful to speak of the "area" beneath the curve, in the Riemann sense (that single, common number is then the assigned area value).
HallsofIvy's example shows a function in which an area in the Riemann sense of the word is impossible to assign to the curve.

ah...ok. so basically, it's saying that IF the upper and lower sums are the same, then we can find the area under teh graph? correct? and that is pretty much the definition of reimann integral right?

 

[arildno]

Yep, as matt grime said, the definition of being Riemann integrable is that the upper and lower sums converge to a single limit.
The fancy word "Riemann integrable" says the same as saying it is meaningful to assign an area beneath the curve (in the Riemann sense of "area").

 

[mathwonk]

riemann integration means trying to squeeze the graph in between some rectangles which lie below the graph, and some other rectangles which lie above the graph, so that the amount of area caught between the two sets of rectangles can be made as small as desired.

one acse whne this is true, i.e. when the function does have a riemann integral, is for an increasing function.

the reason is as follows: subdivide the interval [a,b] say by a point c with a < c < b, and put a rectangle under the part of the graph between a and c, and another one above the graph, and make them fit as close to the graph as possible.

then move over to the interval [c,b] and do the same thing. notice since the graph is increasing that the lower rectangle fopr the right hand interval will be the same height as the upper rectangle for the elft ahnd interval.

hence if you consider the "brick" lying between the upper and lowers rectangles, there is a brick over each interval, but the total height of the bricks, is at most the total height of the whole graph between a and b.

thus if you subdivide into a ,lot of loittle intervals, each of width 1/N, and if the total heoght of the graph is B, then the total area of these bricks, i.e. the difference between the upper and lowers rectangles, wiill be B/N. hence as N gets larger this difference gets smaller.

so the function is riemann integrable.

a picture makes this obvious.

this argument ios due to Newton, and here is a site with hNewtons own illustration of it:

http://www.math.sunysb.edu/~tony/132F02/imagecaption.html

 

[mathwonk]

i have noticed that this site accepts "fazzbazz", but not "oink" as an acceptable post. I find this prejudicial against pigs.

of course it does take "yellow pig", no doubt in deference to m. spivak.

 

[houserichichi]

Zinnnnnnng

 

[Haelfix]

Incidentally, is there a more general theory of integration than integration with say a Lebesgue measure? I'm looking for the cutting edge of mathematical research.

 

[matt grime]

Yes, measure theory is used in integration. I doubt that you'd call it cutting edge though. Measure theory is now more of a tool than a research topic. It's useful in probability, calculus (real and complex), hilbert spaces, banach spaces, dynamical systems...

 

[mathwonk]

the lebesgue definition of the integral is over 100 years old, hence certainly not cutting edge research. It is however a staple of all cutting edge research in analysis, i.e. no one doing cutting edge research uses the riemann version of the integral.

lebesgue's integral is a very simple variant of riemann's, but introduces much mroe technical difficulty.

for example, suppose you wanted to estimate the average incomes of the readers of a magazine.

The Riemann method says take all the readers in each state, and pick one of them at random, and use his income as the estimate for that state.

then do it again, taking only all readers in one county, and again pick one of them at random, and use his income as your estimate for that county.

This is obviously an incredibly inefficient method, that only works well if all people living near each other have similar incomes. i.e. some requirements of continuity are built into riemann's definition.

the lebesgue method says instead, pick a range of incomes and ask readers to check off whether their income lies in the range fro 10K to 20K, or 20K to 30K, etc...

Then you add up all the people who answer 10k-20k, and use say 15k as an estimate for all their incomes.

this is much better, but requires you to be able to count how many people check each block.


so for a function, lebesgue says to subdivide, not the x interval, but the y axis, into subintervals. then for each subinterval of values, you measure how "many" points x of the domain interval exist such that f(x) is in the given interval.

I.e. you try to measure the "size" or "length" of the inverse image under the function, of an interval of the y axis. this is technically quite hard and leads to the theory of measure.

nonehteless, its superiority leads eventually to a better theory, which assigns in some way an integral to essentially all functions.


E.g. an example given earlier was the function which equals 0 at all rationals in [0,1] and 1 at all irrationals. riemann's method simply never converges to an answer for this function, because all nearby points are treated alike by his estimates, whereas there exist nearby points with radically different values.

Under lebesgues method, one says the integral of this function equals 1 times the size of the irrationals, plus zero times the size of the rationals in [0,1].

now since essentially all the points in [0,1] are irrational the integral is 1.

I.e. given any interval of length e>0 no matter how small, one can chop it up into an infinite sequence of disjoint pieces, of lengths e/2, e/4, e/8, e/16,... and still cover up all the rationals in [0,1] using these pieces, since the rationals are countable.

so the rationals have length zero. "hence" the irrationals in [0,1] must reasonably have length 1.

 

[GCT]

Trying to recall from what I learned last week. Basically the graph of the region in which we wish to take the (riemann) integral has to be continuous over the region.

You start by observing a partion P, represented by the interval x - x_{i-1} . The area of the rectangle at a point x_i^* which is between the interval is represented by A^n = f(x_i^*)( \Delta x) . The function itself as well as the \Delta x can be expressed in terms of n where n basically relates to its position in \sum_{i=1}^n . So the Riemann sum, R represents the sum of all of the rectangles with variable partitions as well as the variability in regard to the position of x_i^* within the intervals. R= \sum_{i=1}^n f(x_i^*)( \Delta x) . Now place all of this in terms of n and take the limit as n infinity. So the definition of the integral in relation to the Riemann sum is \lim_{x\rightarrow n} R as R is defined above.

 

[mathwonk]

the correct theorem on existence of riemanns integrals is the following:

a function f defined on a closed bounded interval [a,b], has a Riemann integral over that interval if and only if the set of discontinuities of f has measure zero, (in the sense I demonstrated above that the rationals have measure zero).

In particular you see you cannot even discuss the problem of riemann integrability definitively without the concept of lebesgue measure, or at least of measure zero.

It has always been a pet peeve of mine, for stduents to come away from a calculus thinking that a function needs to be continuous in order to be Riemann integrable.

I.e. since the theory of the integral is designed to strengthen the idea of area, one would surely want to be able to integrate a function whose graph was a series of rectangles, and hence had a finite number of jump discontinuities.

However from the way we teach them, i.e. by saying that all continuous functions are integrable, they get exactly this impression, as perhaps general Chem tutor has done.

for that reason i always emphasize the result i proved above that all monotone fucntuions are integrable, whether continulus or not.

also all bounded functions with only a finite set of discontinuities are integrable.

 

[fourier jr]

Incidentally, is there a more general theory of integration than integration with say a Lebesgue measure? I'm looking for the cutting edge of mathematical research.

godel proved that there is no theory of integration which can deal with every possible function, so i guess there's a limit (in a manner of speaking) on how general a theory of integration can be. there's also the daniell integral & the henstock-kurzweil integral (aka generalised riemann integral).

 

[Haelfix]

Thats cool, I was not aware of the HK-integral, it looks interesting I'll read up on it.

My knowledge of integration theory more or less stops at Lebesgue-Stieltjes types (well I have a working knowledge of other notions of integration in different settings, like say using the Haar measure.. I guess that counts) after undergrad analysis.

Is the HK integral more or less the most general version that you can think off?

 

[mathwonk]

i do not believe any of these integrals works on a larger class of real valued functions than lebegue's.

i think the extra generality is a bit illusory, i.e. one abstracts the properties of an integral in them.

in fact even to produce a function which is not lebesgue integrable as I recall, you need to use the axiom of choice.

i am not an expert either though, and am just thinking out loud here from ancient memory.

haar measure is also not more general then lebesgue integration, or than lebesgue measure. rather the opposite, it is more special. as you surely know, it is merely the name for an ordinary measure which is translation invariant under a given group operation.

let me clarify my use of the words " more general".

if you mean a definition of integration which works on a space with less structure than euclidean space, then these other definitions are more general. but if you mean by more general, as i do, a definition which works on a larger class of functions on euclidean space than lebesgue measure does, then they are not more general.

i.e. in the usual setting these other definitions do not give anything new. they are merely abstractions of the usual definition, which are designed to make sense in a larger class of spaces.

for example, the unique "haar measure" on eucliden space (if normalized by assigning measure 1 to a unit cube, and for the usual group operation of translation), is lebesgue measure.

the daniell integral is treated on page 132-3, of the classic text by riesz nagy, published over 50 years ago, indicating again its traditional nature as a theory of integration. (The original papers of daniell and frechet cited there appeared in 1915 and 1917.) the treatment moreover there is only 2 pages, since the arguments are the same as they have given for lebesgue measure.

briefly, the "daniell integral" consists of defining an integral as a positive linear functional, i.e. something that assigns numbers to functions, so as to be linear, and to take "positive" (in the european sense) functions to positive numbers.

this is nice, but not a big deal in comparison to lebesgue's innovation.

 

[mathwonk]

my impression from frequenting this skte is that the questioners are mainly people who use the web to do their learning and the people who answer the questions (correctly) are mainly people who read books to learn.

I.e. I think it is very hard to actually learn anything on the web alone.

e.g. if you want to know whjat lebesgue meaure means, you might happen upon the following site:

http://planetmath.org/encyclopedia/Integral2.html

where you find an almost useless explanation.

of course it may be that with persistence you will also find somewhere else an explanation of lebesgue measure, the only deep ingredient of this explanation, and the one which is missing here.

OK, I admit that two links further away I found the definition of the key concept of measurability and the definition of lebesgue outer measure.

(say...isn't this version of the theory due to caratheodory? who is not credited.)

 

[fourier jr]

in fact even to produce a function which is not lebesgue integrable as I recall, you need to use the axiom of choice.

it's vitali's example of a non-measureable set that requires the axiom of choice.

Is the HK integral more or less the most general version that you can think off?

as far as i know it is. the HK integral can do both riemann & lebesgue integrable functions. it's weird that it's not taught in 1st-year calculus instead of the riemann integral because it's not a lot more complicated. robert bartle says its definition is "riemann-like", but its power is "super lebesgue". the function f(x) = x^2cos(\pi/x^2) isn't lebesgue integrable on say [0,1] because it's not absolutely continuous, but it's hk-integrable anyway.

Thats cool, I was not aware of the HK-integral, it looks interesting I'll read up on it...

bartle explains it all in "return to the riemann integral" in amer math monthly vol 103 # 8 (oct 1996) pp 625-632

 

[matt grime]

Why isn't that function lebesgue integrable? I mean, what on Earth has absolute continuity got to do with lebesgue integrals? And why say can do Riemann and lebesgue, when being lebesgue integrable means it is Riemann integrable?

 

[Haelfix]

Let f(t) = t^2cos(1/t^2) and f(0) = 0

Well there's a strong pole around t = 0 in that equation, so the Riemann integral looks like this

int (0..1) f'(t) dt.. Which is infinite, the Lebesgue integral doesn't exist, but the improper Riemann integral exists. Namely if you take

lim (epsilon --> 0) int (epsilon..1) f'(t) dt

I guess the HK integral is strong enough to subsume this.

 

[matt grime]

If the improper Riemann integral exists then the lebesgue integral exists. Pole? Where? The area certainly isn't infinite, if it exists, being, as it is, somewhat less than 1 in absolute value. So, what am I missing here?

 

[Haelfix]

Do we agree that the function is not absolutely convergent?

Now consider the simpler example I can think off,

int (0..inf) sint/t dt

clearly this exists as an improper riemann integral
but it is not Lebesgue integrable b/c the integral of the positive part and the integral of negative part are both infinite. Yes?

I'd have to think a little to prove that a Lebesgue integral requires absolute convergence in the series, its usually taken as part of the definition afaik.

 

[mathwonk]

haelfix, you seem to have a very fuzzy grasp of the ideas and terms you are throwing around so dogmatically.

for one very simple and precise case, absolute continuity is a criterion for a function to BE an integral of another function, not to be integrable. this is taught in basic first year real analysis.

as for your example of a function, x^2 cos(<pi>/x^2), that trivial example is obviously not only continuous on [0,1], hence both Riemann and lebesgue integrable there, but even differentiable there.

you are confusing the fact that this function is not itself the riemann integral of its derivative, (this function is not lipschitz continuous, because its derivative is unbounded), with the fact that it is however riemann integrable, hence integrable in virtually every possible sense.

(to be the integral of a riemann integrable function, it is nec to be lipschitz continuous, a stronger version of absolute continuity.)

 

[matt grime]

I'm going to stop replying to you Haelfix since you obviously don't have a clue as to what any of the terms you're using (with great certainty) are. In particular you don't appear to have the slightest idea of what lebesgue measure theory is.

 

[master_coda]

sin(x)/x is not Lebesgue integrable on (0,infinity) because |sin(x)/x| is not Lebesgue integrable on that interval - it diverges. However the improper Riemann integral does exist (I believe it's equal to pi/2).

This is the standard example of a function which is not Lebesgue integrable that can still be integrated using an improper Riemann integral.

 

[mathwonk]

I am also rapidly becoming disenchanted with this series of exchanges. But I think I have identified one of the sources of confusion plaguing Haelfix. Thus I post once more. As usual the confusion is based in a disagreement as to the meaning of words used differently by different people.

I.e. the word "integrable". one meaning is to use this word to distinguish between whether the integral of a given function, assuming it is defined, is infinite or finite.

i.e. one takes the decomposition of the function f into positive and negative parts

f = g-h, where both g and h are non negative, one considers the integrals of g and h separately, and one subtracts these two integrals. of course if both integrals are infinite there is a problem and one usually says the integral is undefined. (this is apparently the situation in the example cited by master coda.) the same trivial distinction occurs between the divergence of the harmonic series and the alternating series obtained from it by changing every other sign.

[to be of interest, the interpretation of this integral as finite has to allow the somewhat unlikely phenomenon, that it has nothing to do with area under the graph (which is infinite).]

Although it has great practical importance whether an integral is finite or not, this is not (to me at least) a deep question in the theory of integration.

I am more interested in the subtle problem of dealing with discontinuites of functions in the theory of integration, as the concept of measurable functions does in lebesgue theory. hence it interests me more if someone has a definition that differs from lebesgue's for positive functions.

Nonetheless the primary distinction between the integrals that have grasped your interest, i.e. between the so called HK integral and the lebesgue integral, is their treatment of integration of unbounded functions. so let's look at the HK integral a moment.

The modern definition, is to me rathjer unintuitive, exactly the opposite of the opinion of its adherents. But anyway here is the history of this "cutting edge" concept:

As a quick websearch, and a little homework reveals, the so called HK integral is merely a reformulation in simpler terms of a definition of the integral based on antiderivatives, that was actually first studied by Denjoy in 1912, and Perron in 1914.

Moreover the idea was suggested in the first edition of the original book of lebesgue!

all this is explained in the book of riesz nagy on page 103 i believe. moreover it is pointed out there that for bounded functions, or even for function with a certain boundedness condition on their derived numbers, there is no difference at all in these definitions and the lebesgue definition.

moreover as explained even by one of the present day exponents of the so called HK theory, this theory is peculiar to the real line, and generalized poorly to other situations, although papers have been written on that topic, by Kurzweil.

i admit however that i have only a little famimliarity with this ap[proach to the integral, which for unboudned functions allows one to assigna number to some of them whose absolute values are not integrable via lebesgue.

so please go ahead, you might well enjoy studying it. may i suggest however, and this to myself as well, one generally receives a better reception here from experts, or even semi experts, if one pretends a little modesty, especially before posting statements rife with errors and misconceptions.

by the way, the only innovation in the HK integral, it seems, is to make the definition of the Denjoy Perron integral resemble at least formally that of the riemann integral. people who think learners are benefited by formal resemblences rather than conceptual ones (I am not one of these people) believe this would make it a good substitute in beginning courses.

In this spirit, MacShane has also given a riemann type definition of the lebesgue integral. it was then proved that a function f such that |f| is HK integrable, is also macshane integrable, i.e. apparently lebesgue integrable.

what this seems to say is, that if you think an integral should not be finite unless the area under the graph is finite, then none of these definitions do anything at all to generalize the lebesgue definition.

for this material see rocky mountain journal of mathematics, vol 34, no. 4, wtr 2004, page 1353., paper by lee tuo yeong.

note however, even in the unbounded case covered by the "HK" definition, there is apparently not one single idea in any of this that was not introduced by lebesgue himself, only reformulations of his ideas.

 

[Haelfix]

Umm, not to be rude but I think this is a terminology thing, not the other way around. If I am brief and slightly inprecise in my posts, its not b/c I don't know better, but rather I do not have a blackboard in front of me and I have been to lazy to get latex working on these boards and do not wish to go in pedant mode for several pages of definitions.

So I have pulled out the text where I first learned measure theory eg Rudin, just to make sure we are all on the same pages, I am more or less copying straight from the text.

Do we agree on the following definition:

Let f be a measurable real valued function, and E is some set, and there is a measure mu.

Consider the two integrals:

int (over set E) max(f,0) dmu , int (over set E) (- min (f,0)) dmu

then the integral
int (over set E) f dmu = difference of the above two integrals respectively.

If both integrals are finite, we say f is integrable on E in the Lebesgue sense with respect to the measure mu

Using this definition I think its clear that say sinx/x from zero to infinity is NOT integrable in the Lebesgue sense.

I suppose you could redefine things and talk of the improper Lebesgue integral, but that's not how I learned things.

(more coming)

 

[Haelfix]

Oh I just saw MasterWonk's post, yea we are more or less talking about the same thing.

Rather than talking from a text not everyone has (eg Rudin), I just googled for a few seconds and found a course paper with some of the relevant definitions
http://www-finmath.uchicago.edu/Courses/Stochastic2.pdf

Look on page 6, they use absolute convergence (not absolute continuity, I never said anything about that) in the definition of what is or is not integrable.

(Incidentally,being a theoretical physicist, we use the word pole somewhat informally and sloppily - mea culpa. Regardless in the writeup of the HK integral that function that Fourier Jr suggested is mentioned in several places, it was a motivating concept behind their work. Moreover, please keep in mind the f(0) = 0 condition)

-edited to remove potential ambiguites in decorum =)

 

[Hurkyl]

Here's a problem from Royden, third edition, page 93, 10b:

The improper Riemann integral of a funciton may exist without the function being integrable (in the sense of Lesbegue), e.g., if f(x) = (sin x)/x on [0, &infin;]. ...

 

[mathwonk]

haelfix,

i apologize for accusing you of the error involving absolute continuity. that one was in Fourier jr's post.

perhaps i was too hard on you for your error calling a nice differentiable point of a function a "strong pole". to me a pole is a point where the value of an analytic function is infinity (look for example in rudin where you say you learned the subject), whereas this function had value zero at this point. I fail to see how there could be any difference of opinion about such a matter.

question: when i give you useful and correct historical information from a standard famous source, why do you fault me for using a source "not everyone has"?

when i reveal to you that ideas you think are "cutting edge" mathematical research are 90 years old, why are you unhappy to learn this?

this is not a debate, this is an attempt to arrive at understanding. some of us here are actually mathematical scholars, and we use appropriate source material. if this were merely a contest to see who could top the other person we might limit our sources sportingly to the american mathematical monthly.

if we were discussing theoretical physics which you say is your specialty, i would be glad to learn from you about it.

have a nice day.

 

[master_coda]

Look on page 6, they use absolute convergence (not absolute continuity, I never said anything about that) in the definition of what is or is not integrable.

They use absolute convergence when referring to the integrability of a simple function. Which makes some sense, since the Lebesgue integral of a simple function is defined with a summation, and you want that summation to converge absolutely.

But saying that integrability requires absolute convergence makes no sense when referring to an arbitrary function. Absolute convergence is a property of a series, not of a function.

 

[mathwonk]

let me try to say this this as simply as possible (limited by my knowledge and understanding).

MY impression is the following:

if you think "integral" should mean area under the graph, then there is no definition more general than lebesgues.

i.e. to integrate a positive function, there is no better or newer approach than that of lebesgue, all later approaches agree with his, and none work on a larger class of functions.

to integrate a function with both positive and negative values, there are various definitions yielding different results.

they all agree when both positive and negative parts have finite integrals, but when both positive and negative parts have infinite integrals, then there are various different ways to decide how to combine these integrals.

1) lebesgue's method apparently says to look separately at the positive and negative parts, integrate them separately and subtract.

in particular if f has finite lebesgue integral, then so does |f|.
this is the most restrictive convention.

2) other methods such as improper riemann and HK integration allow one to cancel the positive and negative parts against each other so that some functions whose positive and negative parts both have infinite integrals, yield when combined, a finite integral.

If this interests you, you may be interested in the HK integral.

nonetheless, whenever |f| has a finite HK intergal, then it already has a finite lebesgue integral, so nothing is gained from the HK approach, as proved in the paper i cited, which is available on the web.

If this is in error, please correct me.

 

[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, &infin;)... just have the n-th approximation be zero on [n, &infin;).

 

[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.