Wikipedia:Reference desk/Archives/Mathematics/2009 April 15

= April 15 =

Composition of functions and integrability
Is there an example of a discontinuous integrable f and a continuous integrable g such that f(g(x)) is non-integrable? I know the converse is not true - if f is continuous and integrable, and g is integrable, then f(g(x)) is integrable, but what about with g continuous? I'm fairly confident there is no counterexample, but I'm not totally sure how to prove it - how would I go about it if that is the case?

Thanks very much,

Mathmos6 (talk) 13:38, 15 April 2009 (UTC)


 * The following homeomorphism g should help you to build a counterexample with f Riemann integrable and $$\scriptstyle f\circ g$$ not even measurable in Lebesgue sense. Let h(x):=x+c(x) where c : [0,1] → [0,1] is the Cantor function; then h : [0,1] → [0,2] is a homeomorphism (for it is continuous & strictly increasing) taking the Cantor set C into a closed subset A of [0,2] of measure 1 (for the complement of C in [0,1] is sent into a set with the same measure, that is 1). Therefore g:=h-1 : [0,2] → [0,1] is a homeomorphism taking A into the Cantor set. Notice that, since A has positive measure, it contains a non-measurable set (you don't need this, if you are just happy with $$\scriptstyle f\circ g$$ not Riemann integrable). Can you see how to go ahead? --pma (talk) 14:42, 15 April 2009 (UTC)

Well the standard example for f-integrable g-integrable with composition fg not integrable is g Thomae's function, f(x)=1 everywhere except f(0)=0 - so is the next step something similar to that? I can't honestly say I'm completely sure how to go ahead, despite the fact you've already given me a lot of help - sorry, my head's obviously having a slow night!

Mathmos6 (talk) 19:06, 15 April 2009 (UTC)
 * The function you describe is (Lebesgue) integrable. Are you interested in Riemann integrability? Algebraist 19:10, 15 April 2009 (UTC)


 * OK, start with an example with f Riemann integrable, g a homeomorphism (the one defined above) and with composition $$\scriptstyle f\circ g$$ not Riemann integrable. Choose f to be a characteristic function, $$\scriptstyle f:=\chi_S$$. Then $$\textstyle \chi_S \circ g =\chi_{g^{-1}(S)} $$ is also a characteristic function. Thus you want S such that $$\scriptstyle \chi_S $$ is Riemann integrable and $$\scriptstyle \chi_{g^{-1}(S)}$$ is not. To this end you just have to clarify a bit to yourself which sets S have a Riemann integrable characteristic function (these are called Jordan measurable sets; if you are familiar with the characterization of Riemann integrable functions by means of their points of continuity it's quite immediate). Then, if you recall the relevant property of the above homeomorphism g, you are done. But we still do not know if you are interested in Riemann or in Lebesgue integrability. In the latter case, if you want a stronger example, you can in fact choose a Jordan measurable set S such that the set $$\scriptstyle {g^{-1}(S)}$$ is not even Lebesgue measurable (in this case use the little remark above). --pma (talk) 21:05, 15 April 2009 (UTC)

That's brilliant! I understand completely I think, except for one thing - what's the relevance of h being homeomorphic? Is that just so we know it's continuous, or does it have additional relevance? As a matter of interest, is it safe to 'adjust' the example so that g is [0,1]->[0,1], say by simply multiplying h(x) by (1/2)? Thanks so much for the help, Mathmos6 (talk) 04:46, 16 April 2009 (UTC)


 * You're welcome. Of course,  h(x)/2 is even nicer, for the interval remains the same. The fact that h is a homeomorphism has no particular relevance for your needs, except that being an additional property, makes the counterexample stronger. Moreover, it shows that such properties of sets, like: having zero measure, or: being measurable either in the Lebesgue or in the Jordan sense, are not topological invariants.
 * Note also: the measurability problem may be bypassed if one chooses to deal with functions f that are measurable in the sense of Borel, which is a topological concept, rather than in the sense of Lebesgue, which is not (the subtle difference is that in the former case the sublevel sets {f<c} are required to be Borel measurable; in the latter, only Lebesgue measurable). After all, any Lebesgue measurable function is (canonically) equal a.e. to a Borel measurable function, so you would lose nothing in terms of classes of functions defined a.e. For Borel measurable f, of course, it is true that $$\scriptstyle f\circ g$$ is still Borel (with g continuous, or even Borel measurable). BUT the problem essentially remains, in the sense that the map $$\scriptstyle f\mapsto f\circ g $$, although now well-defined in a suitable class of measurable functions, is not well-behaved: still changing f in a set of zero measure could change $$\scriptstyle f\circ g $$ in a set of positive measure.
 * On the positive side, if g is a homeomorphism and g-1 is Lipschitz, then $$\scriptstyle f\circ g $$ is Riemann (or Lebesgue) integrable if such is f; the reason is that now g-1 can not expand the null sets where f is bad-behaved to fat ones. In this case, the map $$\scriptstyle f\mapsto f\circ g $$  gives rise to a nice linear continuous operator between the corresponding L1 spaces. --pma (talk) 08:55, 16 April 2009 (UTC)

Isolated singularities
does poles belong to an isolated singularity?? —Preceding unsigned comment added by 59.96.30.227 (talk) 15:08, 15 April 2009 (UTC)


 * Yes, a pole is a special case of an isolated singularity. — Emil J. 15:36, 15 April 2009 (UTC)

Algebra: Olympiad question
This is an Olympiad question which I cannot think of any way to start. "The +ve integers a and b are such that 15a + 16b and 16a - 15b are both squares of +ve integers. What is the least possible value that can be taken by the smaller of these squares." Please help. --Siddhant (talk) 16:54, 15 April 2009 (UTC)
 * This is quite easy, as IMO questions go. The answer is 231361. Algebraist 17:27, 15 April 2009 (UTC)

How? I know the answer but I need the method. Thanks for your effort.--Siddhant (talk) 17:32, 15 April 2009 (UTC)
 * Show that both squares are divisible by 481. Algebraist 17:53, 15 April 2009 (UTC)

Do I need to use Fermat's theorem for that?--Siddhant (talk) 18:44, 15 April 2009 (UTC)
 * I didn't use any theorem of Fermat. That doesn't mean there's not an approach that would use such a theorem, however. Algebraist 18:46, 15 April 2009 (UTC)
 * Well, Siddhant, have you proved the answer? If you still have not got it, you can always ask again and we will certainly help you. The trick with these problems is not to think too hard because usually the solution only requires a few basic facts from number theory (and minor algebraic manipulations). -- PS T  02:27, 17 April 2009 (UTC)

I assumed let r2=15a + 16b and let s2=16a - 15b. I squred both the squares to get r4 + s4 = 481 (a2 + b2). Now how to prove that each of the squares is divisible by 481? What arguments need to be given after that to reach the answer? Is there another approach possible to answer this question? Thanks.--Siddhant (talk) 08:06, 17 April 2009 (UTC)
 * Write a and b in terms of r2 and s2 and then consider what happens if 481 doesn't divide one of the squares. Zain Ebrahim (talk) 10:19, 17 April 2009 (UTC)
 * You want to show that r and s are both divisible by 481, and you have that r4 + s4 is divisible by 481. Divisibility by 481 is the same thing as divisibility by 13 and 37. Suppose s is not divisible by 13: then, working modulo 13, you have (r/s)4=-1. Why is this impossible? Alternatively, give up on this and try a different approach, such as Zain's. Algebraist 11:35, 17 April 2009 (UTC)
 * If you are doing IMO, you presumably work with a professor. In that case, it would be best to ask him/her considering you have been trying at this for some time. -- PS T  12:41, 17 April 2009 (UTC)

Not the OP, but I've gotten sucked in myself. I can show that r (and/or s) is divisible by 481 (thanks to the above hints), and so get that r^2 must be a multiple of 231361, but how do I know that's actually a solution to the original equations? Don't I need to find workable values of a and b? 74.7.169.90 (talk) 21:58, 17 April 2009 (UTC)
 * Yes, but that's not hard. Algebraist 22:16, 17 April 2009 (UTC)