Convergence of lebesgue integral

In summary, The function fn(x) is given by n times the characteristic function of the interval (0, n^-1) minus n times the characteristic function of the interval (-n^-1, 0), for x between -1 and 1. It is required to prove that the limit as n approaches infinity of the integral of fn(x) with respect to x is equal to the integral of the function f(x). This can be done using direct computation or the dominated convergence theorem. However, in order for the dominated convergence theorem to be applicable, we need to find a function g(x) that dominates all the f_n's, which is not possible because the absolute value of f_n is not dominated by any function in L1.
  • #1
onthetopo
35
0

Homework Statement


[tex]fn(x)=n\chi_{(0,n^{-1})}(x)-n\chi_{(-n^{-1},0)}(x)[/tex] -1<x<1
Prove that [tex]\int fn(x)dx\rightarrow\int f(x)[/tex]
a. Using direct computation
b. Use dominated convergence

The Attempt at a Solution


a. How to do direct computation?

b. Using dominated convergence
If x is in (0,1/n), then fn(x)=<1-0=1
If x is in (-1/n,0), then fn(x)=<0-1=-1
If x is in neither , then fn(x)=0
Thus |fn(x)| is dominated by |g| where g(x)=1 , a constant function on (-1,1)
g(x) is in L1, since g(x) is Riemann integreable on (-1,1) it must also be Lebesgue integreable.
Thus, by dominated convergence theorem:
"limit of integration of fn" is equal to "integration of limit of fn"= integration of f(x)=0 as n goes to +infinity.
Thus,

[tex]\int fn(x)dx\rightarrow\int f(x)\rightarrow0 [/tex] QED
 
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  • #2
[itex]\int f_n[/itex] isn't that hard to comptue... What have you tried?
 
  • #3
Can we say that
[tex]\int fn(x)=\int n\chi_{(0,n^{-1})}(x) +\int -n\chi_{(-n^{-1},0)}(x)[/tex] -1<x<1 (**)
I think the above is the crucial step. What exactly do we need for the above to hold? I guess as long as each term on the right is integreable?

Assume (**) holds, then let u be the lebesgue measure
[tex]\int n\chi_{(0,n^{-1})}(x) +\int -n\chi_{(-n^{-1},0)}(x)=n*u(0,n^{-1})-n*u(-n^{-1},0)=n*1/n-n*(1/n)=0[/tex]

Is my answer to part (a) posted here , together with my answer to part (b) posted in the first post,both completely correct? =)
 
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  • #4
Can someone please tell me if what i did is right?
 
  • #5
Yup, that's good.

Your answer for part (b) is incorrect however. f_n isn't dominated by 1. Try to sketch f_1, f_2, and f_3 to get a feel of what's going on.
 
  • #6
If x is in (0,1/n), then fn(x)<n*1/n-0=1
If x is in (-1/n,0), then fn(x)>0-n*1/n=-1
If x is in neither , then fn(x)=0
Thus |fn(x)| is dominated by |g| where g(x)=1, I really can't see what's wrong?
 
  • #7
No, if x is in (0,1/n), then f_n(x)=n; and if x is in (-1/n,0), then f_n(x)=-n.
 
  • #8
ah!
so it's dominated by |n|, which is not in L1, so we can;t use DCT at all
 
  • #9
What do you mean?

What you want to do is find a nice integrable function g on [-1,1] that dominates all the f_n's. It doesn't look like you've tried to do this.
 
  • #10
if x is in (0,1/n), then f_n(x)=n; and if x is in (-1/n,0), then f_n(x)=-n., and if x is in neither, f_n(x)=0
thus, |f_n(x)|<|n| for all x on (-1,1) , but|n| is unbounded as n goes to infinity
we cannot find a g(x) in L1 such that |fn|<|g| for all n and all x in (-1,1).
 
  • #11
That doesn't mean you can't fit a g in between f_n and n. I don't immediately see a good candidate though; and I'm willing to guess that you're right, dct probably won't work. You can use a generalized version of the dct (say, the one mentioned in your other thread), but that's kind of stupid though, because it amounts to basically doing what you did for part (a).
 

FAQ: Convergence of lebesgue integral

1. What is the Lebesgue integral?

The Lebesgue integral is a mathematical concept used to calculate the area under a curve. It was developed by French mathematician Henri Lebesgue in the early 20th century as a more general and powerful alternative to the older Riemann integral.

2. How does the Lebesgue integral differ from the Riemann integral?

The main difference between the Lebesgue and Riemann integrals lies in the way they partition the interval of integration. The Riemann integral divides the interval into equal subintervals, while the Lebesgue integral divides it into smaller segments based on the behavior of the function being integrated. This allows the Lebesgue integral to handle a wider range of functions, including discontinuous and unbounded ones.

3. What is the importance of the convergence of Lebesgue integral?

The convergence of Lebesgue integral is crucial in ensuring that the integral is well-defined and yields a meaningful result. Without proper convergence, the integral may be undefined or give inconsistent results. Therefore, understanding the different types of convergence and how to apply them is essential in effectively using the Lebesgue integral.

4. What are the different types of convergence in Lebesgue integral?

There are three main types of convergence in Lebesgue integral: pointwise convergence, almost everywhere convergence, and convergence in measure. Pointwise convergence means that the function being integrated converges to a limit at every point in the interval of integration. Almost everywhere convergence means that the function converges to a limit at all points except for a set of measure zero. Convergence in measure is a weaker form of convergence, where the function converges to a limit in a probabilistic sense.

5. Why is the Lebesgue integral preferred over the Riemann integral?

The Lebesgue integral has several advantages over the Riemann integral, making it the preferred method of integration in many applications. It can handle a wider range of functions, including those that are discontinuous or unbounded. It also allows for easier and more general methods of integration, such as the use of Fubini's theorem. Additionally, the Lebesgue integral has a more intuitive geometric interpretation, making it easier to understand and apply in practical situations.

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