Lebesgue Integral

  • #1
is this a lebesgue intergral? And why?

[tex]\int\frac{dx}{x^2}[/tex]

P.S Can u also give me exact defintion of Lebesgue integral PLz. Textbook tooooooo complicated.
 
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Answers and Replies

  • #2
matt grime
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At the moment it isn't an integral. You need to put in a dx or something. it is a lebesgue integral if you're integrating using the measure theoretic definition for integral. however, sidestepping the issue of the function not being defined at zero and that there is no limit in the integral, it is riemann integrable, as it's a continuous function.

remember to end environments with "/" in brackets, not "\".

Do you really need to learn measure theory? I can think of other things you need to learn before you get that far that appear to be missing (cf your other thread) before starting an advanced topic such as measure theory.
 
  • #3
Ok. forgive my mathematical incompetency Mr Matt Grime :tongue: . I have to do learn this coz its part of General integration theory.
 
  • #4
One question mATT! how do u know all this stuff? Are u a lecturer or masters student.??????????????????????????????
 
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  • #5
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Hi:

My answer to you is yes !

But i can't understand your meening when you ask "why".
So please explain what you know already about measure theory.

Moshek

Try to be carfull from Math answers...
 
  • #6
matt grime
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I am not a lecturer nor am I a masters student. I am somewhere in between.

I'm just surprised that they're making you do this stuff given you were new to the idea of proving that other result about |f| by contradiction. Please don't take offence, I am genuinely bemused by that sylllabus.

As it is my knowledge of measure theory is very ropey so I won't begin to explain the intricate aspects of it. Something is a legesbue integral if it use lebesgue integration, you don't indicate in that example what kind of integral you are doing, but it is lebesgue integrable (since it is riemann integrable).
often something like du(x) is used to indicate one needs to use lebesgue techniques.

One way of doing the lebesgue integral is, I believe, to use step functions to approximate the function in some way.

Say we have f defined on the unit interval to be 1 at irrationals and 0 at rationals. Then that is not Riemann integrable. But it is lebesgue integrable because it is equal to the constant function f(x)=1 except on a set of measure zero, hence its integral is the same as the integral of the constant function, ie 1. (Someone set me straight if I'm mucking it up).
 
  • #7
I know thats riemann integrable coz its bounded and has no points of discontinuity over that interval. riemann integrable implies lebesgue integrable. is that a good enough reason for why its lebesgue integrable?
 
  • #8
hmm. matt,I wasnt totally new to the idea of proving by contradiction, its just that I'm not good with mathematical proofs and this lecturer really is something when he marks stuff. If i leave anything no matter how trivial it is he subtracts marks. That kind of question is intuitively obvious but I suck at putting it into mathsy type o language.
 
  • #9
matt grime
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well, assuming the interval over which your integrating is indeed as you say it is then that's fine.
 
  • #10
The interval is [1,infinity)
 
  • #11
matt grime
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then that isn't fine as the interval is unbounded. the same observations you made apply to 1/x and that isn't integrable over that region. (and i made a mistake too in the reply.)
 
  • #12
Oh interval has to be closed. ure right. is it an improper riemann integral??
 
  • #13
matt grime
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absolutely, that integral exists.
 
  • #14
[tex]\[
\int_1^\infty {\frac{{dx}}{{x^2 }}} = 1
\]
[/tex]

Just in case you were wondering.
 
  • #15
thanx!!i figured that part cos
that integral
= 1 + lim 1/n (as n tends to infinity)
= 1

just trying to figure out if our questions are lebesgue integral,riemann integral or improper riemann integral and why??
So this one should be a improper riemann integral,,,rite???
 
  • #16
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0
A function f is Lebesgue integrable if it is measurable and int(|f|) is finite. However just because a function is bounded and measurable does not imply that it's Lebesgue integrable. For example the constant function 1 is not Lebesgue integrable on [1, infinity). However, for your problem, you could use the fact that it's both positive and Riemann integrable.
Vignon .S Oussa
 
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