Seminar on Measure Theory 2

This is the second in a three part lecture course on measure theory and integration. This lecture will be an introduction to the Lebesgue integral, following on from Seminar on Measure Theory 1 in December (which you should read first). The only prerequisites are those of the previous talk plus the contents of the previous talk itself. It would also be useful to know what lim sup and lim inf are.

The contents of the talk include the motivation and definition of the Lebesgue integral, the monotone convergence theorem and proofs of basic results such as linearity of the integral.

The talk transcript

21:55 (@cyby`) Shortly, we will be having the second part of Kit's seminar on Measure Theory. Stay tuned!
21:57 Kit tests his microphone
21:57 ( notjanet) microphone??
21:58 ( dmhouse) it's working then.
21:58 ( notjanet) audio too?
21:58 (@Kit) Wow that's loud. People all the way over in freenode can hear me.
21:58 (@Kit) Now excuse me while I get rid of a cat who has other ideas about me giving this talk, and then I suppose I'll be ready to begin.
21:59 ( dmhouse) notjanet, a metaphorical microphone.
21:59 ( dmhouse) i guess.
21:59 (@Kit) Everything is metaphorical.
21:59 (@Kit) You will note that I am also not talking, but merely typing on a keyboard.
22:00 (@Kit) Still no truth in advertising. :(
22:00 ( dmhouse) :)
22:00 ( snel_hest) worked for me
22:00 (@Kit) Ok. I suppose it's about time to begin.
22:01 (@Kit) As per usual, if you have a question say !
22:01 (@Kit) Then ask it when I get to you.
22:01 (@Kit) Allow me to demonstrate
22:01 (@Kit) !
22:01 (@Kit) Kit: Yes?
22:01 (@Kit) Please don't interrupt me mid train of thought, 'cause I'll get all flustered and stuff. :)
22:02 ( notjanet) ..
22:02 ( Syzygy-) !
22:02 (@Kit) Syzygy-: Yes?
22:02 ( Syzygy-) Will you ignore our !s until your train of thought has finished so that you would avoid being flustered?
22:02 (@Kit) Yes. :)
22:02 (@Kit) The train of thought takes frequent rest stops.
22:03 (@Kit) So, integration.
22:03 (@Kit) When I asked Brouwer to give a talk on operator algebras his response was "Why would you want me to do such a thing? It would consist entirely of slander and swearing."
22:03 (@Kit) This struck me as a good idea.
22:03 (@Kit) So I'm going to begin with the talk with a few lines of slander.
22:03 (@Kit) The Riemann integral sucks.
22:03 (@Kit) Why?
22:04 (@Kit) It's not because we've got silly functions which aren't Riemann integrable.
22:04 (@Kit) So the characteristic function of Q isn't Riemann integrable. Who really cares?
22:04 (@Kit) All we really desperately need are the continuous functions.
22:04 (@Kit) The problem with the Riemann integral is that it has bad theorems.
22:05 (@Kit) For example, the continuous functions. The first analysis course I attended didn't prove that continuous functions are Riemann integrable.
22:05 (@Kit) It needs ideas of uniform continuity they thought were too advanced.
22:05 (@Kit) Proving things to be Riemann integrable is quite a lot of work.
22:05 (@Kit) Secondly, exchange of limits.
22:06 (@Kit) Some of you have probably seen my magic 'introduce a parameter, fiddle with it, the integral drops out because of swapping orders' tricks for calculating definite integrals.
22:06 (@Kit) Why does this work?
22:07 (@Kit) One of the first lessons you learn in analysis is that it's not always valid to swap limits around.
22:07 (@Kit) Lets take a basic example. Suppose we have functions f_n, f with f_n(x) -> f(x) for every x
22:07 (@Kit) If the f_n are integrable, is f neccesarily? And if so does int f_n -> int f?
22:07 (@Kit) Well the answer is no in general, as some easy examples show. But what conditions make it true?
22:08 (@Kit) This is very hard to answer in the case of the Riemann integral.
22:08 (@Kit) So, we instead have the Lebesgue integral which is all round friendler. :)
22:08 (@Kit) It gets its power from the fact that we did all taht work to get measure spaces.
22:08 (@Kit) Once you have a measure space you can do integration on it, and the measure does all your work for you.
22:09 (@Kit) What we will do is essentially define the integral on a small class of functions, and progressively increase the functions we have it defined for.
22:09 (@Kit) Until eventually we have it defined for a sufficiently large class of functions to make us all happy, and then we'll prove some strong limit theorems about it.
22:10 (@Kit) So, now for some definitions.
22:10 (@Kit) Let X be a measure space. A function f : X -> R is said to be measurable if for every Borel subset U of R, the set f^{-1}(U) is in the sigma-algebra of X.
22:10 (@Kit) Note I'm not assuming f to be invertible.
22:10 (@Kit) f^{-1}(U) means { x : f(x) in U }
22:10 (@Kit) What motivates this definition?
22:11 (@Kit) Well, basically when we're integrating things we'll want to look at things like m({ x : a <= f(x) <= b })
22:11 (@Kit) And we want this to make sense, so we'll need the set { x : a <= f(x) <= f(b) } to be in the sigma-algebra.
22:11 (@Kit) And this is precisely the condition that f^{-1}([a, b]) is measurable.
22:12 (@Kit) Easy check: If f^{-1}(A) is measurable for all A in some collection which generates the Borel sigma-algebra then f is measurable.
22:13 (@Kit) In particular it will be useful to note that the sets of the form [a, infty] generate it.
22:13 (@Kit) Oh. Damn. I should have warned you.
22:13 (@Kit) I don't actually mean R there.
22:13 (@Kit) I mean [-infty, infty]
22:13 ( zeno) I think you meant <= b in that second set, not <= f(b)
22:13 (@Kit) Yes, sorry.
22:13 (@Kit) But see comment about the care and feeding of ! :)
22:13 ( Reloaded) [-infty, infty] ?
22:14 (@Kit) See comment about see comment about...
22:14 (@Kit) And I'm about to explain anyway.
22:14 ( Reloaded) Oh, OK.
22:14 ( Reloaded) (I'm just not sure how an interval can be closed at infinity)
22:14 (@Kit) [-infty, infty] is what we call the extended reals. It's all the real numbers, together with -infty and infty (infty is what I shorten infinity to)
22:14 (@Kit) These are just arbitrary symbols we introduce for convenience.
22:14 (@Kit) They're ordered in the obvious way
22:14 (@Kit) And we can also do arithmetic on them.
22:15 (@Kit) For example x + infty = infty for any real number x
22:15 (@Kit) We don't allow infty - infty
22:15 (@Kit) We do however allow 0 infty and declare it to be 0.
22:15 ( Reloaded) Oh really..
22:15 (@Kit) Oh shush.
22:15 (@Kit) I'll put +m on and devoice you if you don't.
22:16 ( Reloaded) ...
22:16 (@Kit) Anyone who wants voice can ask in #math
22:16 (@Kit) My lovely assistants will help.
22:17 (@Kit) Now, train of thought lost. Please bear with me.
22:17 (@Kit) Right. Borel measure can be put on the extended reals just by extending it so that the singletons {infty} and {-infty} have measure 0
22:17 (@Kit) But this doesn't matter especially. We only care about the sigma algebra.
22:18 (@Kit) Anyway, the measurable functions will be the ones for which we define the Lebesgue integral.
22:18 (@Kit) We'll need a useful lemma:
22:18 (@Kit) Let f_n be measurable functions, then lim sup f_n is measurable.
22:18 (@Kit) We'll first show that sup f_n is measurable.
22:19 (@Kit) Sorry, these operations are pointwise.
22:19 (@Kit) i.e. (lim sup f_n) (x) = lim sup (f_n(x))
22:19 (@Kit) Anyway, as I observed the sets [a, infty] generate the Borel sigma-algebra, so its' enough to show that if f = sup f_n then f^{-1}([a, infty]) is measurable for each a
22:20 (@Kit) Sorry. I want (a, infty]
22:21 (@Kit) But f^{-1}((a, infty]) = { x : sup f_n(x) > a } = { x : for some n f_n(x) > a } = U f_n^{-1}(a, infty)
22:21 (@Kit) And the sigma algebra is closed under countable unions, so this set is measurable as each of the f_n is
22:21 (@Kit) We can do essentially the same to show that inf is
22:21 (@Kit) But then lim sup f_n = inf_m sup_{n >= m} f_n
22:22 (@Kit) So lim sup is measurable.
22:22 (@Kit) Now, this is important because it means that if f_n -> f pointwise and the f_n are measurable then so is f.
22:22 (@Kit) I'll use this a lot
22:23 (@Kit) So we've got a decent chance of proving good limit theorems.
22:23 (@Kit) Now, on to integration.
22:23 (@Kit) What we will do is start with a certain class of functions: The characteristic functions of measurable sets.
22:23 (@Kit) The characteristic function of a set A is I_A(x) = 1 if x is in A, 0 if x is not in A.
22:24 (@Kit) For functions like this we'll define Int I_A = m(A)
22:24 (@Kit) So the integral is in some sense extending the definition of the measure.
22:24 (@Kit) Now, we want integrals to be linear. i.e. if t, s are real numbers then Int(tf + sg) = t Int f + s Int g
22:24 (@Kit) So we can use this to extend the definition to finite linear combinations of the characteristic functions.
22:25 (@Kit) i.e. things of the form sum_1^n z_k I_{A_k}
22:25 (@Kit) We call these the simple functions, and define their integral to be sum z_k m(A_k)
22:25 (@Kit) Exercise: This is well defined.
22:25 (@Kit) Exercise 2: A function is simple iff it's measurable and only takes finitely many values.
22:26 (@Kit) Now, the reason these things will be useful is because every measurable function can be well approximated by simple functions.
22:26 (@Kit) I'm going to define a sequence d_n : [-inf, inf] -> R of simple functions which will be useful in proving this.
22:27 (@Kit) d_n = sum_{k = -4^n}^{4^n} k 2^{-n} I_{ [k 2^{-n}, (k + 1)2^{-n} ) }
22:27 (@Kit) Phew
22:28 (@Kit) Totally incomprehensible probably. :)
22:28 (@Kit) The important part is that d_n(x) -> x pointwise.
22:28 (@Kit) And further it does so in an increasing manner for x >= 0
22:28 (@Kit) And decreasing for x <= 0
22:29 (@Kit) So, let X be some measure space and f : X -> [-inf, inf] be measurable.
22:29 (@Kit) Then d_n f is a sequence of simple functions tending to f
22:29 (@Kit) And further if f >= 0 then d_n f tends to f monotonically.
22:30 (@Kit) This will allow us to prove a whole lot of useful things.
22:30 (@Kit) First of all, if f and g are measurable then so is f + g
22:30 (@Kit) Because (d_n f) + (d_n g) is simple, and so measurable, and we have (d_n f) + (d_n g) -> f + g pointwise.
22:30 (@Kit) And measurable functions are closed under pointwise limits.
22:31 (@Kit) Also any continuous function f : R -> R is measurable, because f d_n is simple (easy check), and f(d_n(x)) -> f(x) because f is continuous and d_n(x) -> x
22:31 (@Kit) Both of these have nicer proofs if you know some topology, but I'm not assuming you do so this is a good easy way to do it.
22:32 (@Kit) Anyway, now we've got some of the legwork out of the way lets finally define the integral.
22:32 (@Kit) At least part of the way.
22:33 (@Kit) Suppose we've got some measurable function f with f >= 0
22:33 (@Kit) We define Int f = sup { Int g : g is simple and g <= f }
22:33 (@Kit) (Again, these inequalities are all pointwise)
22:33 (@Kit) Now, note that we've potentially got two definitions when f is simple
22:34 (@Kit) But this isn't a problem. You can easily verify that for simple functions if f <= g then Int f <= Int g
22:34 (@Kit) Which shows that the definition agrees here.
22:34 (@Kit) (Oh, other easy check: Integration is linear on the simple functions)
22:34 (@Kit) Right. Now that's done, lets prove a theorem. :)
22:34 (@Kit) (Bet you thought I wasn't going to)
22:35 (@Kit) Theorem: The monotone convergence theorem.
22:35 (@Kit) Let f_n : X -> [0, inf]
22:36 (@Kit) If f_n is monotonic increasing (i.e. f_n(x) <= f_m(x) whenever n <= m) and f_n -> f(x) then Int f_n -> Int f
22:37 (@Kit) This is a very important theorem.
22:37 (@Kit) It's basically the machine which will let us prove everything we want to do.
22:37 (@Kit) Unfortunately the proof is a magic trick. :)
22:38 (@Kit) Anyone who has questions should ask for +v in #math and then follow the appropriate procedure.
22:38 (@Kit) I'm not going to respond to pm mid talk.
22:39 (@Kit) Anyway, let c < 1 be arbitrary
22:39 (@Kit) This is like the epsilon trick I used last time. I'm going to make the c go away at the end.
22:40 (@Kit) Anyway, first of all note that we have that integration is monotone. So the sequence Int f_n is monotone increasing, and thus tends to some limit by standard elementary analysis
22:40 (@Kit) (Specifically it tends to Sup Int f_n )
22:40 (@Kit) We'll call this limit q
22:40 (@Kit) So what we want to do is show that q = Int f
22:40 (@Kit) We have f_n <= f, so we know that q <= Int f
22:40 (@Kit) We'll show that q >= c int f
22:41 (@Kit) And then by magic the c will go away, because it was arbitrary.
22:41 (@Kit) Let g <= f be some simple function.
22:41 (@Kit) (Remember that Int f = sup Int g as g ranges over all such simple functions)
22:42 (@Kit) Ok. So let E_n = { x : f_n(x) >= c g(x) }
22:42 (@Kit) Because c < 1 and g(x) <= f(x) we have that for any x it is eventually in some E_n
22:43 (@Kit) Sorry. > c g(x)
22:43 (@Kit) (As f_n(x) -> f(x) > c g(x) unless f(x) = 0 in which case we already have x in E_n )
22:43 (@Kit) No, being stupid. I do want >= after all.
22:44 (@Kit) Anyway, this means that U E_n = X
22:44 (@Kit) Now, Int f_n >= Int f_n I_{E_n}
22:44 (@Kit) (Because f_n >= f_n I_{E_n} )
22:45 (@Kit) And f_n I_{E_n} >= c g I_{E_n}
22:45 (@Kit) Because f_n >= c g on the set E_n, and this function is 0 elsewhere.
22:45 (@Kit) So Int f_n >= c Int g I_{E_n}
22:45 (@Kit) Now we'll show the integral on the right tends to Int g
22:46 (@Kit) This is because we can write g = sum z_k I_{A_k}
22:46 (@Kit) So g I_{E_n} = sum z_k I_{A_k int E_n}
22:46 (@Kit) And A_k int E_n is an increasing sequence of measurable sets with union A_k, so its measure tends to m(A_k)
22:47 (@Kit) So because limits commute with sums we have Int g I_{E_n} = sum z_k m(A_k int E_n) -> sum z_k m(A_k) = Int g
22:47 (@Kit) i.e. Int f_n >= c Int g
22:48 (@Kit) Err.
22:48 (@Kit) I'm being lame.
22:48 (@Kit) Rather lim Int f_n >= c Int g
22:48 (@Kit) (Because we've taken the limit on both sides)
22:48 (@Kit) Now, the c magically goes away and we have lim Int f_n >= Int g
22:49 (@Kit) And then Int f is the supremum of all possible gs on the right hand side
22:49 (@Kit) So we have lim Int f_n >= Int f as desired.
22:49 (@Kit) QED
22:49 (@Kit) I'll open the floor to questions for a moment
22:49 (+yoyoyowat) I have a question.
22:50 (@Kit) Thanks Syzygy-
22:50 (@Kit) Any real questions?
22:51 (@Kit) Anyway, I'll leave the channel -m for now. Feel free to ! if you have questions.
22:51 (@Kit) Ok. This lets us extend a lot of important results which are easy for simple functions.
22:51 (@Kit) For example, let s, t >= 0 be real and f, g >= 0 measurable functions
22:52 (@Kit) Then Int (sf + tg) = s Int f + t Int g
22:52 (@Kit) Why?
22:52 (@Kit) Because it's true for simple functions
22:52 (@Kit) Now take simple functions f_n, g_n tending monotonically to f and g
22:52 (@Kit) Then Int f_n -> Int f and Int g_n -> Int g
22:52 (@Kit) And sf_n + t g_n tends monotonically to sf + tg
22:53 (@Kit) So Int (sf_n + tg_n) -> Int (sf + tg)
22:53 (@Kit) And because the integral is linear for simple functions, we have Int(sf_n + tg_n) = s Int f_n + t Int g_n
22:53 (@Kit) So the two limits are equal
22:53 (@Kit) Right. So we've defined it for positive functions.
22:54 (@Kit) Now we want to extend it to functions f : X -> [-infty, infty]
22:54 (@Kit) Except we can't, quite.
22:54 (@Kit) Basically what we want to do is write f = f_+ - f_-, where f_+ and f_- are both positive functions, then let Int f = Int f_+ - Int f_-
22:54 (@Kit) But this doesn't quite work.
22:55 (@Kit) A detail I've been glossing over is that these integrals can quite happily be infty
22:55 (@Kit) For example the integral of 1 over R is infty.
22:55 (@Kit) This is fine in the positive case.
22:55 (@Kit) But if we tried to do the above then we might get infty - infty
22:55 (@Kit) Which is bad and wrong.
22:56 (@Kit) So what we do is we define a function to be integrable if Int |f| < infty
22:56 (@Kit) And for measurable f we let f+ = max{f, 0} and f- = max{-f, 0}
22:56 (@Kit) Then f+, f- <= |f|
22:56 (@Kit) And f = f+ - f-
22:57 (@Kit) So if Int |f| < infty then we have Int f+, Int f- < infty
22:57 (@Kit) And the above works.
22:58 (@Kit) Oh, I missed a nice theorem.
22:58 (@Kit) Suppose f_n >= 0 are measurable
22:58 (@Kit) Then sum Int f_n = Int sum f_n
22:58 ( Reloaded) if one of f+ and f- has infinite integral and the other has finite then the integral is still defined.
22:58 (@Kit) (I told him about interrupting)
22:59 (@Kit) And yes, he's quite right. This doesn't matter.
22:59 (@Kit) Anyway, so yes. This exchange of sum and integral is an example of how versatile the lebesgue integral is for exchanging limits.
23:00 (@Kit) There's an even nicer example, but unfortunately I don't have time to prove the dominated convergence theorem which it needs.
23:00 (@Kit) So I'm going to stop here.
23:00 (@Kit) Thanks for coming.
23:00 (@Anil) ;)
23:00 ( dmhouse) :) thankyou.
23:01 ( trelaybot) <Anil> thank you very much kit :-)
23:01 (@Kit) Next lecture will be on L^1 as a complete normed space (not as I'd previously advertised on fourier series. I decided that would be too hard at this point)
23:01 (@Kit) It will be on the 28th
23:01 (@Kit) Same time.
23:01 (@Kit) Anil: You're quite welcome. Hope you enjoyed it.
23:01 ( dmhouse) L^1?
23:02 (@Kit) L^1 is another name for the set of integrable functions.
23:02 ( trelaybot) <integral> complete normed space? That's just a normed space, and the topological idea of completeness?
23:02 (@Kit) This will all be explained next time. :)