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Lebesgue integral


Xerxes

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Can anybody help with this? I'm trying to do function spaces, and bumped up against the Lebesgue integral. My books give a pat on the head, and tell me not to worry over much about it, then proceed to tell me it's "zero almost everywhere". That, as a definition, is about as much use as a rubber hammer.

 

I am very familiar with the Reimann integral, but can anybody gentle me through this?

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If you want a concise intro to lebesgue integrals, I'd recommend :

 

Mathematics for Physicists (Dover) by Dennery & Krzywicki,

Chapter III (Function space, orthogonal polynomials and fourier analysis), page 184 -188 ("Elementary Introduction to the Lebesgue Integral")

 

In some searching around that I did a little while ago, when struck by the same problem, I found the following references :

 

1. Strook, D. W. A Concise Introduction to the Theory of Integration, 2nd ed. Boston, MA: Birkhäuser, 1994.

 

2. Henstock, R. The General Theory of Integration. Oxford, England: Clarendon Press, 1991.

 

3. Kestelman, H. Modern Theories of Integration, 2nd rev. ed. New York: Dover, 1960.

 

I have no idea about the quality or usefulness of these last three references. I'd be grateful too, if matt would throw in his suggestions.

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If you want a concise intro to lebesgue integrals' date=' I'd recommend :

 

[i']Mathematics for Physicists [/i] (Dover) by Dennery & Krzywicki,

Chapter III (Function space, orthogonal polynomials and fourier analysis), page 184 -188 ("Elementary Introduction to the Lebesgue Integral")

 

Thanks for that. I have this book, it's one of those I find less than helpful on this subject. On the whole, though, it is a splendid book, and a really good read, thanks again. I especially liked their treatment of analytic functions.

Thanks also for the web sites, but I think I need an interactive session on this, as I am totally baffled.

 

Cheers.

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lebesgue integration is best ignored, as the book states, but let me give you the definition of "almost" that this is talking about.

 

Let R be the real numbers. A set S in R has measure zero if given e>0 there is a collectuion of open intervals that cover S that has total length less than e.

 

example: let S be any coutnable set, and let s_r be the elements in S with r in N, whcihch we can do sinvce S is countable.

 

about the element s_r place an open interval of width e2^{-r} then these discs cover S and their total width we can easily compute is the geometric series

 

e(1/2+1/4+1/8+...)=e

 

so any coutnable set has measure zero.

 

what lesbegue measure does is this: it assigns the size b-a to the interval [a,b) and then extends this measure to all sets that can be obtained by intersecting, complementing, taking the union of these basic sets, and do so additively.

 

 

notice that (a,b) [a,b] and (a,b] all have measure b-a since we are adding or subtracting one point at a time and a point p is contained in the set [p,p+e) and has measure less than e for all e>0 hence it must be zero.

 

the idea then is loosely that to work out an integral of f(x) we look at the points where f(x)=[t,t+e) and work out the measure of this set of x's and assign a contribution of e times the measure (a little like adding up lots of rectangles of area) and work out the limits and see what happens. It is all very messy.

 

the upshot of it is that if we take a fucntion like

 

f(x)=0 if x is rational and 1 if x is irrational, then the integral over the intercval [0,1] doesn't exist in the riemann sense but does in the lesbegue sense. f is only not zero on a set of measure zero (the rationals which are countable) and 1 on the rest, thus the area must be 1.

 

it is usually hard to tell what the measure of a set is, and is very hard to find a set that is not measurable (there are some but they are patholgically bad in a technical sense)

 

the moral of the story is that behaviour on a set of measure zero can be ignored. if we ignore a set of measure zero then we are still saying "almost everywhere" ie that last function is almost everywhere equal to 1 since the the places it isn't 1 are a set of measure zero. if two functions agree everywhere except a set of measure zero (ie almost everywhere) then they have the same measure theoretic features, that is the integrals are the same over any given region.

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Thanks for the "feeler" matt. This is what I shall settle for until I have the need to revisit this in greater detail (ie: actually have to solve Lebesgue integrals)

 

I had "almost" convinced myself to walk the following path : topological spaces -> sigma-algebras -> delta-rings, measures -> measure spaces, riemann, lebesgue measures -> riemann, lebesgue integrals.

 

PS : the "=" following the sum of the widths of the discs should be a "<"

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nope, think that the sum of the gp 1/2+1/4+1/8+... is 1. there are two other = signs but they don't need changing either.

 

i really do'nt know anything about lebesgue integration; this is all stuff i've cobbled together f from various sources that feels like a reasonable idea of what it is. i've had to use measures for probability but i never really understood them except to konw that it can safely be ignored. it is a useful tool where, once you develop a certain level of intuition, what yuo want to be true is true.

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nope, think that the sum of the gp 1/2+1/4+1/8+... is 1.
Don't know what I was thinking...ignore that.

 

The wall I found myself running against was in understanding completeness (for QM). I got a feel for completeness (of an infinite set of orthogonal functions) in terms of being able to write any continuous function as a linear combination of the basis functions. Beyond that, one runs into Lebesgue integrals :-(

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the moral of the story is that behaviour on a set of measure zero can be ignored. if we ignore a set of measure zero then we are still saying "almost everywhere" ie that last function is almost everywhere equal to 1 since the the places it isn't 1 are a set of measure zero. if two functions agree everywhere except a set of measure zero (ie almost everywhere) then they have the same measure theoretic features, that is the integrals are the same over any given region.
Well, well. Got it now, nice explanation. Thanks. It also explains why, if the Reimann integral exists, it is equal to the Lebesgue. Neat.
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