Re: need help understanding Lebesgue integration

On 20 Apr, 19:29, jaialai.technol...@xxxxxxxxx wrote:
As an exercise I need to calculate the Lebesgue integral of a function
over a certain interval.

Can I please get some advice(hints) on how to Integrate(using only the
definition of the lebesgue integral!)
This function f on the interval [-3,3]?
f(x)= (1 + signum (sin pi/x))

signum (sin pi/x))  = -1 for x >-1
signum (sin pi/x))  = 1 for x >1

So, from [-3,-1] f(x)=0
from [1,3] f(x)=2
on [-1,1] f(x) gets pretty wacky though!

In any event, I am not even sure how to apply the definition of the
lebesgue integral to this problem...

IIRC, the definition of the Lebesgue integral I've seen is the
supremum of the integral of all simple functions <= f(x). First,
consider the sequence of simple functions f_n(x), where

{ 0 for x <= -1
f_1(x) = { 2 for x >= 1
{ 0 otherwise

and f_{n + 1}(x) is given by f_n(x) + g_n(x), where

{ 2 for x in (1/(2n + 1),1/(2n))
g_n(x) = { 2 for x in (-1/(2n - 1),-1/(2n))
{ 0 otherwise.

The integral of f_n is given by 4 + sum_{i = 1}^{n - 1} 4/(4i^2 - 1),
so the integral of f is >= 4 + sum_{i = 1}^oo 4/(4i^2 - 1) = 6 (I
think).

It remains to be shown that, if F is any simple function <= f, then
the integral of F is less than or equal to 6 (so that the supremum of
the set of integrals of all such F's is indeed given by sup int f_n).
To show this, note that the simple function F_{n + 1} given by

{ f_n for x > 1/(2n + 1)
F_{n + 1} = { f_n for x < -1/(2n - 1)
{ 2 otherwise

is >= f, and so any F has an integral <= that of F_{n + 1}. Also

int f <= int F_{n + 1} = int f_n + 8n/(4n^2 + 1) <= 6 + 8n/(4n^2 - 1)..

Since this is true for any n, and since the second term -> 0 as n ->
oo, it follows that int f = 6.

(I haven't checked the arithmetic all that carefully, but an argument
along these lines should give the correct answer if what I've written
is wrong).
.

Relevant Pages

  • Re: Numerical methods for Lebesgue integrals?
    ... As I'm just learning about Lebesgue, ... integrals which are not Riemann-integrable evaluate to 0. ... there would be no real difference between Riemann and Lebesgue: ...
    (sci.math.num-analysis)
  • Re: proof: xf(x) -> 0
    ... Riemann, Lebesgue or something else? ... when you are using either the Lebesgue or improper Riemann integral. ... be defined as the supremum of all the integrals of simple functions g ... that depend on the precise definition: the Monotone Convergence Theorem ...
    (sci.math)
  • Re: Numerical methods for Lebesgue integrals?
    ... those Lebesgue integrals for which Riemann doesn't exist? ... Even when Riemann exists, so Lebesgue = Riemann, could numerical techniques ...
    (sci.math.num-analysis)
  • Re: complexity of integrals...
    ... Lebesgue integration the use of higher order language appears ... than the Lebesgue integrals? ... This makes the Riemann extension "elementary" in the ... Herman Rubin, Department of Statistics, Purdue University ...
    (sci.math)
  • Re: need help understanding Lebesgue integration
    ... lebesgue integral to this problem... ... I didn't prove that this sum is correct, since the proof I had in mind ... the set of integrals of all such F's is indeed given by sup int f_n). ...
    (sci.math)