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Measure Theory Discussion
By Brad Rodgers on Monday, September 03, 2001 - 11:09 pm:

Suppose that a function is defined such that

f(x)=0 for x¹0
f(x)=¥ for x=0

Would this imply that ò-1 1f(x) dx=c for c¹0?

I could intuitively see this going either way. On one hand, it is very similar to Dirac's delta function. On the other hand, I could imagine taking the function at x=0, splitting it up into countably infinite segments, and then rearranging it into what would seem to have an area of zero...

Brad

By Dave Sheridan on Tuesday, September 04, 2001 - 03:46 pm:

Repeat after me, "Infinity is not a number."

You can't define f(x) to be infinity if f is going to be a function - or if you do, you'd better make sure you know exactly how to deal with infinity in every way you need to (for example, how does integration behave for infinite functions?)

So the simple answer is that you haven't defined a function and it's pretty much because of this type of problem that we can't just define functions to be infinite at any point. In fact, the Dirac delta function is not a function either, for precisely this reason. It's a measure.

A measure is a way to associate a "length" with every subset of the real line. Actually, there's much more to it than that and you don't need to just work with the reals, but that's the idea. Lebesgue measure is the one you know something about, saying the measure of a set (a,b) is b-a. This is what happens when you use a ruler to measure a line, for example. Integration is continuous because the measure of a single point is always zero.

The Dirac measure assigns length 1 to any set continaing 0, and length 0 to all other sets. This leads to integration being discontinuous at zero, and that is precisely why the Dirac "function" isn't defined at zero.

The function you were trying to describe is just c×Dirac measure. It can't be described in terms of normal (Lebesgue) integration because Dirac measure is sufficiently different from Lebesgue measure that the two can't be described in terms of each other. I'll go further into this if you like.

Let me know if this all makes sense and I'll write some more about how to integrate with respect to a measure, rather than a variable...

-Dave

By Brad Rodgers on Wednesday, September 05, 2001 - 01:58 am:

Alright, sorry about the ill defined nature of my 'function' (could we call it a distribution- like the delta function). I know we went through this once, and I think I understood it. The problem here was, I didn't know of a better way to express myself (seeing as how I think you got the idea, would there be a rigorous and non-contradictory way to express that 'function'?)

Anyway, I'm relatively illiterate of the delta function, knowing only its definition and some of the basic properties (relation with the heavside function, it's use for defining functions with integrals, etc.)

Could you tell me a bit more about Lesbegue Integration. I tried reading a bit on this, only to find myself overwhelmed (partially because the document I was reading was overridden with errors). I did manage to learn a bit on measure though, and I already new quite a bit about groups and sets. The problem I encountered was that while I knew what infimum and supremum were defined as, I hadn't really dealt with these before and couldn't really appreciate the defintions very much either, so wasn't very apt at using them. Could you give a reference to a site about these?

I would also like to hear more about integrating with respect to a measure.

Thanks,

Brad

By Dave Sheridan on Wednesday, September 05, 2001 - 01:53 pm:

I'm afraid I don't know where decent reference sites are, and most of the books around are very difficult since they're aimed at people who already have an undergraduate degree in Maths... Having said that, the book of the Cambridge measure theory course is pretty good. It covers measure theory from a probabilistic viewpoint, which isn't everyone's cup of tea but is much easier to understand than most heavy-going textbooks. It's called "Probability with Martingales" by David Williams, published by CUP.

For inf and sup, you should think of them as max and min for infinite sets. So if I look at the set of 1/n, there is no minimum value. However, as n tends to infinity you see they approach zero from above. We'd like to say the minimum of the set is zero, but that isn't strictly true (zero is never attained). So we say the inf is zero. It's a bit like looking at the minimum of the closure of the set, if that helps. Similarly with sup/max.

Ok, so integration - what does it all mean? Lebesgue integration is quite difficult to understand in general, so I'll go from the bottom up.

Suppose we have a measure (this simply assigns a "length" to each set we supply to it. This can't be done on ×every× subset of the reals consistently in the way we like, so in general we have to restrict the sets we can measure to what's known as a sigma field, but for the purposes we'll deal with, pretty much every set you'll come across is measurable. You need the axiom of choice to come up with a non-measurable set and even then you can't explicitly construct one easily). I'll call the measure m, so that m(A) is the length of A.

Recall, Lebesgue measure behaves as follows:
m([a,b])=m((a,b))=m([a,b))=m((a,b])=b-a,
and this has to behave consistently. By consistent, I mean that the measure of a disjoint countable union of sets is the sum of the measure of each of the sets and the empty set has measure zero. So if you can construct a set from intervals (open, closed or whatever) using countably many set operations, you can measure it. This is pretty much how you get sigma algebras, by the way...

So, what is an integral with respect to m? Suppose we have a function f(x)=1 if x is in A, and f(x)=0 otherwise, and A is a measurable set. The integral of this function should be m(A), so that's what we define it to be. Essentially, we're integrating a constant over the set A.

Now we expand this by allowing linear combinations of these functions, and defining the integral to be a linear operator. Finally, if you have a monotonically increasing sequence of functions (ie f_n(x)<=f_{n+1}(x) for all x), the integral of the limit is the limit of the integrals. So again, all we're doing is being consistent with those simple functions, knowing that integration is linear.

Perhaps surprisingly, this is enough to know that the integral of x is x2/2, and so on. But only when we're dealing with Lebesgue measure.

How should we write this? Well, there are various ways to do it. Some people write integral f dm, some people write m(f) and some people write integral f(x) m(dx). All mean the same thing. When we deal with probabilities, the measure is normall y denoted P, so P(A) is the measure of a set A, and we call this the probability of the event A. When X is a function (we call this random variable) the integral of X is normally denoted E(X), or expected value of X. This is to be consistent with basic probability theory, but allows us to use measure theory.

Ok, so what about the delta function? Clearly the integral of f(x) mentioned above is 1 if A contains 0, and 0 if A doesn't. If we extend this linearly, all we end up doing is picking out f(0). So the integral of any function g with respect to the delta function is just g(0).

Probabilistically, this corresponds to saying that with probability 1, 0 is chosen. Imagine you've got a roulette wheel which is fixed so that the ball always ends up at 0. The distribution of the location of the ball is a delta function located at zero. We call this an atom at zero, of mass 1.

If I wanted to record the outcome of the toss of a coin, I need an atom at "H" (I could write X=0 to mean a head) of mass 1/2, and an atom at "T" (or X=1 to mean a tail) of mass 1/2. So I could say that the measure associated with a coin is 1/2 times the delta function, plus 1/2 times the delta function shifted up the number line by 1. I've added two measures here. The integral of the sum is the sum of the integrals, so if I integrate a function g against this, I'll end up with 1/2 times g(0) plus 1/2 times g(1), which is exactly what we need - with probability 1/2, we record what happens if we have a tail, and with probability 1/2 we record what happens if we have a head.

Let me know if I've gone too quickly and I'll go over this again; otherwise I'll tell you how you can combine the discrete case (delta functions) with the continuous case (Lebesgue integration) and indeed give you some examples of random variables which correspond in some way to Lebesgue integration.

-Dave

By Arun Iyer on Wednesday, September 05, 2001 - 07:45 pm:

Wow!! That's quite a long message...phew!!

Anyway BRAD!!
Try this site for some info on Lebesgue integration!!
lebesgue integration

love arun

By Brad Rodgers on Friday, September 07, 2001 - 02:42 pm:

I think I understand what you wrote, however, I'm not really sure how to apply this to areas of functions. I do think I've understood what you've written thus far though.

Thanks,

Brad

By Dave Sheridan on Friday, September 07, 2001 - 04:29 pm:

To see how integration relates to areas of functions, we first have to understand what we mean by "area". We say that the area of a rectangle is its length times its width. This implicitly requires an underlying measure or we wouldn't understand what "length" or "width" meant.

When you first encounter integration, you chop a function up into small x increments and approximate the shape by boxes, whose area you can work out explicitly. If the approximation gets arbitrarily good as the size of the increments goes to zero, the function is integrable and the limit of the areas is the integral.

What if you chopped the function up into y incrememnts instead? So, consider the set of points x such that f(x) is in the region [y,y+dy]. You could do exactly the same thing as above, but conceptually you're rearranging the function too, because you're lumping everything at the same height together. So one way to do it is to reorder your function so that it's decreasing, ie start at 0 with the highest point and go down from there. If the function is continuous, this won't affect the area but it means you can work things out from first principles like you did with the x bits. However, you don't need to work it out from first principles if you have measure theory behind you. We can find the "length" of the set of x such that f(x) is in [y,y+dy] since we already have a measure. That's essentially what the integration does.

For a more concrete way of looking at it, consider the function which I mentioned above - it's 1 if x is in the set A, and 0 if x isn't in A. First, consider A=(a,b). This is a simple step function then, starts at 0, jumps to 1 just after a, and jumps back to 0 at b. What's it's area? Length times height. The height is clearly 1, and the length is b-a. Whoops! No, the length is m((a,b)) which is b-a under Lebesgue measure. But under (say) the Delta measure the length is 0 if a>0. Similarly, you can generalise so that the "area" of the indicator of a general set A should always be m(A). This happens to be consistent, so we use it.

The most used example of a function which is Lebesgue integrable but not Riemann integrable (the previous definition of integration) is the function which is 1 if x is rational and 0 if x is irrational. The upper integral has a limit of 1 and the lower integral has a limit of 0 so it's not Riemann integrable. But the Lebesgue measure of the set of rationals is 0 (trust me on this for the while) so the Lebesgue integral is simply 0. In effect, we've squashed all of the 1s of the function together and measured their length, which happens to be 0. So the integral is 1×0 + 0×length of the irrationals.

Does this make sense? If so, we can progress further. If not, let me know and I'll explain it slightly differently.

-Dave

By Brad Rodgers on Friday, September 07, 2001 - 08:33 pm:

I think I understand this thus far. Is the fact that the measure of rational numbers is 0 related to the countablitity of the rationals?

By the way, what is the proof that there are a countable number of rationals? I've heard this before, and I've seen a proof for the uncountablility of the reals, but have never actually been able to find a proof for rationals.

Thanks,

Brad

By Olof Sisask on Saturday, September 08, 2001 - 12:18 pm:

You can list the rationals <1 in order of their denominators (excluding repetitions such as 2/4 and 3/6 for 1/2):


1 2 3 4 5 6 ...
0/1 1/2 1/3 2/3 1/4 3/4 ...


Now, every rational number can be made up of one of these fractions plus an integer, and since the result of adding one countable sequence to another is still countable, the rationals are countable.

Regards,
Olof
By Dave Sheridan on Monday, September 10, 2001 - 11:53 am:

Another way to do this is simply to specify an injection such as p/q -> 2p × 3q. This handles positive rationals and negative ones follow if -p/q is injected to (say) 5p × 7q.

Yes, the fact that the rationals have zero measure is precisely because they're countable. If A is the countable union of disjoint sets then the measure of A is the sum of the measure of each of the sets. This is part of the definition of measure. The rationals are the countable union of each rational number, and the measure of any particular number is 0 since it is [a,a] for some a. Note that this is not true for Dirac measure! Now the countable sum of 0 is still 0. Why? Because it's the limit as n tends to infinity of 0+0+...+0, which is always 0. If you try to add uncountably many 0s together, this does not hold (think about integrating 0 to obtain 1).

-Dave

By Brad Rodgers on Tuesday, September 11, 2001 - 01:11 am:

What would be the area under a function f(x),

f(x)={1 if x irrational, but not trancendental
.......{0 if x trancendental or rational

By David Loeffler on Tuesday, September 11, 2001 - 12:44 pm:

0. The set of algebraic numbers is countable, so it has zero measure; hence your function is 0 almost everywhere.

Proof: An algebraic number is the root of a polynomial with integer coefficients. For the polynomial Sn i=0aixi define its "height" to be Sn i=0(1+|ai|).
Then, since each ai is an integer, the height is a +ve integer; and there are clearly finitely many polynomials having a given height, each with finitely many roots, we see that the set of algebraic numbers must be countable.

(I seem to recall that Cantor discovered this before anybody had succeeded in providing a single example of a transcendental number!)

[Editor: One intuitive way of understanding why there are so many more irrationals (uncountable) than rationals (countable) is to consider pressing the digits 0,1,..,9 randomly on your calculator to generate a decimal expansion. The resulting number is rational precisely when the expansion eventually repeats itself ad infinitum. What's the chance of that!]