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 Posted: Tue Jul 24, 2007 6:56 pm
Question from my assignment Quote: It can be shown that every interval on the real line contains both rational and irrational numbers. Accepting this to be so, do you believe the function f(x)= 1 if x is rational 0 if x is irrational is intergrable on the closed interval [0,1]? Explain your reasoning?
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Posted: Tue Jul 24, 2007 7:37 pm
Fortunately for you, this case doesn't make a big distinction between Riemann-Stieltjes integrability and Lebesgue integrability.
However, you do need Lebesque integrals and Lebesgue measure theory to deal with this problem.
So before I answer anything, which class is this for?
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Posted: Wed Jul 25, 2007 2:28 pm
For my Calculus class (level one at uni). We just started integration a couple of weeks ago.
I had a brainstorm last night (while I was sleeping)
I'm thinking now that it is not integrable on the closed interval, because it is a peicewise function, but it can't be broken down into different intervals because it say that every interval has both rational and irrational numbers.
Does that sound right?
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Posted: Wed Jul 25, 2007 6:39 pm
Soy un hombre muy honrado,
que me gusta lo mejor My thinking is that on any interval, there will be a jump discontinuity. Las mujeres no me faltan,
ni el dinero ni el amor
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Posted: Thu Jul 26, 2007 3:10 pm
There is a jump discontinuity on every interval. In fact, there is a jump discontinuity everywhere.
The question then becomes "is this actually a problem?" After all, you can integrate over some jump discontinuities, and if it's a removable discontinuity then you can ignore it altogether. Sure, the answer seems to be that you simply integrate on either side of the discontinuity, but is that really the only solution?
The issue is actually slightly more subtle than it seems. Consider the number of the set of rational numbers between 0 and 1, and compare that to the number of irrational numbers between 0 and 1. How many removable discontinuities can you ignore?
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Posted: Thu Jul 26, 2007 10:41 pm
Wow, I think you are making it way more complicated than it is.
Though I do see your point.
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Posted: Thu Jul 26, 2007 11:28 pm
Dewdew Wow, I think you are making it way more complicated than it is. Though I do see your point. It depends. The answer is simple; after all, it is a yes or no question. The reasoning behind the answer is actually rather complicated, because while ordinary calculus can't do the "ignore an infinite number of discontinuities" trick, that doesn't mean that the answer has to be no. The size of that infinite number of discontinuities matters a lot here; in fact, the entire question hinges on it. So, even though you don't have the tools to answer this question, I ask you, suppose you ignored all of the rational numbers between 0 and 1, so that you're integrating a function on just the irrational numbers. How much area are you losing from the integral by doing so?
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Posted: Sat Jul 28, 2007 10:49 pm
I was surfing wiki a moment ago and found this:
http://en.wikipedia.org/wiki/Riemann_integral#Examples
Warning, it's the entire answer explained step by step, so if you want to reason it out on your own, this is major spoilers.
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Posted: Sun Jul 29, 2007 2:01 am
CodeLabMaster I was surfing wiki a moment ago and found this: http://en.wikipedia.org/wiki/Riemann_integral#Examples Warning, it's the entire answer explained step by step, so if you want to reason it out on your own, this is major spoilers. More spoilers (kinda): The entire problem is that calculus on its own doesn't distinguish between the two types of integrability. So the question of whether something is integrable isn't actually well-defined.
For instance, in this case the area beneath the function can be figured out after a bit of thought, but whether an integral would return that value depends on the type of integral. Calculus, however, only teaches a version of the Riemann integral.
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Posted: Thu Aug 02, 2007 10:49 pm
Just for the record I think it's supposed to be Riemann integration.
I should get the answrs on Monday so I can post the "correct" answer, or at least according to the marker
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