- 15 Jul '07 20:14 / 2 editsokay, so this is the third and final question in my real (and abstract!) analysis exam this year. part i) is just a definition, part ii) requires a wise choice of partition and part iii) still evades me. and i got a good mark in the rest. it is just part iii). i wrote maybe a line, but i doubt i got any marks for it . so, can someone help me see part iii)? all of them need quite a good knowledge of the analysis branch of pure maths (specifically, integration).

so,

3. Let a, b be in the set R with a= R be a bounded function. Explain what it means for f to be integrable, and define the integral of f.

ii) Let f,q:[a,b] -> R be functions and let E be a finite subset of [a,b]. prove that f is integrable and q(x)=f(x) for all x in the set [a,b]\E, then q is integrable and int(q(t)dt,b,a) = int(f(t)dt,b,a).

iii) Let q:[a,b] -> R be a real function. Assume that for each n in the set N, the set (E subscript n)={x in the set[a,b] | |q(x)|>=(1/n)} is finite. Show that q is integrable and compute the integral int(q(t)dt,b,a).

[HINT: For each n in the set N consider the function (q subscript n):[a,b] -> R defined by (q subscript n)=

{0 if x is in the set [a,b]\(E subscript n)

{q(x) if x is in the set (E subscript n)

for x in the set [a,b]. Then use part (ii).]

anyone?

*note:*int(f(t)dt,b,a) would be the integral of f(t) wrt t between b and a (with b at the top and a at the bottom).

*noteII:*any reference to integration is referring to the riemann integral... - 17 Jul '07 10:22

Can you please upload the exam?*Originally posted by genius***okay, so this is the third and final question in my real (and abstract!) analysis exam this year. part i) is just a definition, part ii) requires a wise choice of partition and part iii) still evades me. and i got a good mark in the rest. it is just part iii). i wrote maybe a line, but i doubt i got any marks for it . so, can someone help me see part iii)? ...[text shortened]... ottom).**

*noteII:*any reference to integration is referring to the riemann integral...

The part where you define E_{n} is quite hard to understand in the way you write it. - 17 Jul '07 12:42This is clearer using Lebesque integrals. Each set E_n is a zero measure set. E is the union of all the E_n's and is also a zero measure set (it's countable). So the integral must exist and be zero since [a,b]/E contains all values of x for which q(x) < (1/n) for all n in N.

In the context of the Riemann integral it's less straightforward. You can probably define functions q_n(x) which are equal to q(x) for all x in [a,b]/E union E_n, and zero in E/E_n. Each of these is integrable (since the set E_n is finite points in E_n are isolated) and the integral is zero. q(x) is the sum of the q_n's which is also zero. - 17 Jul '07 16:00I wanted to edit my last post but the site went down before I could. I hadn't read the question properly and misunderstood the hint so my "you can probably define" should be replaced with "as they say in the hint, define" - I was having problems with your notation, not your fault - we should petition Russ for a [math] tag for equations.

I think I've got the basic method, but to make it acceptable to mathematicians you need to say things like "for any epsilon you can find n st 1/n < epsilon" to demonstrate that q is zero in the set [a,b]/E. - 17 Jul '07 16:25

I doubt it'll be high on his list of priorities, but if you want to propose it then I'll support it. Code to do this stuff exists so it probably isn't that hard to implement, but it's also the kind of thing that's quite time consuming to test and so on.*Originally posted by adam warlock***Yes we should. Something like latex would be very handy. Should I take it to site ideas?** - 17 Jul '07 16:29

And if you consider the amount of tags that already exist, introducing math tags would more than double the existing ones.*Originally posted by DeepThought***I doubt it'll be high on his list of priorities, but if you want to propose it then I'll support it. Code to do this stuff exists so it probably isn't that hard to implement, but it's also the kind of thing that's quite time consuming to test and so on.** - 17 Jul '07 16:50

Already post it. Let's see how it goes.*Originally posted by DeepThought***I doubt it'll be high on his list of priorities, but if you want to propose it then I'll support it. Code to do this stuff exists so it probably isn't that hard to implement, but it's also the kind of thing that's quite time consuming to test and so on.** - 17 Jul '07 17:39

i can't upload the exam as i lack a scanner and my lecturer hasn't put it up yet (and probably won't until next april). you could try writing the question down?*Originally posted by DeepThought***This is clearer using Lebesque integrals. Each set E_n is a zero measure set. E is the union of all the E_n's and is also a zero measure set (it's countable). So the integral must exist and be zero since [a,b]/E contains all values of x for which q(x) < (1/n) for all n in N.**

In the context of the Riemann integral it's less straightforward. You can ...[text shortened]... _n are isolated) and the integral is zero. q(x) is the sum of the q_n's which is also zero.

we haven't touched on the lebesque integral-it is covered in a masters course which i -erm- shan't be taking... also, the riemann integrals are a subset of the lebesque integrals. that is, if someone is riemann integrable => lebesque integrable, but not . or so i have been led to believe. - 17 Jul '07 23:29

Since I've answered the question I can't be bothered to write it down. The Lebesque integral is defined in terms of measures of sets, with the measure of the interval [0,1] defined as 1. If a function is Riemann integrable it is Lebesque integrable, but the converse is not necessarily true. The standard example is Dirichlet's function which is 1 on the rationals and 0 on the irrationals. It's Riemann integral over any interval is undefined - and there is no procedure like the above to make it work, but the Lebesque integral is zero since the rationals are a countable set and the measure of a countable set is zero.*Originally posted by genius***i can't upload the exam as i lack a scanner and my lecturer hasn't put it up yet (and probably won't until next april). you could try writing the question down?**

we haven't touched on the lebesque integral-it is covered in a masters course which i -erm- shan't be taking... also, the riemann integrals are a subset of the lebesque integrals. that is, if someo ...[text shortened]... e is riemann integrable => lebesque integrable, but not . or so i have been led to believe. - 18 Jul '07 07:27 / 1 edit

i was telling adamwarlock to write it down-sorry if it sounded like i was telling you too...*Originally posted by DeepThought***Since I've answered the question I can't be bothered to write it down. The Lebesque integral is defined in terms of measures of sets, with the measure of the interval [0,1] defined as 1. If a function is Riemann integrable it is Lebesque integrable, but the converse is not necessarily true. The standard example is Dirichlet's function which is 1 on th is zero since the rationals are a countable set and the measure of a countable set is zero.**

(also, we've touched on it wrt our lecturer showing us how bad the riemann integral is and telling us that the lebesque integral works for this function when the riemann doesn't. i think theat was the Dirichlets's function-the name rings a bell. he also carried out an experiment-yes, and experiment! in a pure maths class!-where we counted coins. and, surprise surprise, we counted all the 10ps first, then the 20ps then the 50ps etc., which is (alledgedly) what the lebesque does, while the riemann integral just counts them as it gets them...or something. it was a course that confused me just as much as it fascinated me!) - 18 Jul '07 12:37 / 2 edits

No worries. It occurred to me after that there's another way of doing the problem that doesn't especially depend on how you define your integrals or use the hint.*Originally posted by genius***i was telling adamwarlock to write it down-sorry if it sounded like i was telling you too...**

(also, we've touched on it wrt our lecturer showing us how bad the riemann integral is and telling us that the lebesque integral works for this function when the riemann doesn't. i think theat was the Dirichlets's function-the name rings a bell. he also carried ...[text shortened]... gets them...or something. it was a course that confused me just as much as it fascinated me!)

E_1 is a subset of E_2 is a subset of E_3 etc. E_n is finite for all n. I didn't realize this earlier, it means that in the above proof you don't sum the integrals over q_n you just notice that they are zero for all n in N, which completes the proof.

E_n union a union b defines a set of M(n) open intervals which we'll label with m.

The function is bounded above by 1/n and below by -1/n within these intervals which means that the integral of q(x) in one of these intervals is bounded by L_m /n above and -L_m/n below (where L_m is the length of the interval). The sum of these integrals is then bounded above by (b-a)/n and below by (a-b)/n. Given some epsilon you can always find n such that (b-a)/n is smaller than epsilon. So the integral is zero.

Edit: Although there is a problem with all this. The Dirac delta function is defined to be zero everywhere except at x=0 and it's integral is 1. So E_1 = E_2 = ... = {0}. Using either proof we end up with its integral being zero which it isn't. The only resolution to this that I can think of is that delta is not a map from R --> R everywhere, anyone got any ideas?