Problem related to signed measure

  • Thread starter Thread starter onthetopo
  • Start date Start date
  • Tags Tags
    Measure
Click For Summary

Homework Help Overview

The problem involves a measure space (X, S, u) and a function f in L1, with the goal of demonstrating the existence of a set E with finite measure such that the difference between the integrals over E and X is less than a given epsilon.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to define a function v(A) as the integral of f over set A, suggesting that v(A) behaves like a signed measure. Some participants question the necessity of sigma-finiteness in the context of the problem, while others express uncertainty about the correctness of the initial approach.

Discussion Status

The discussion is ongoing, with participants exploring the implications of sigma-finiteness and its relevance to the problem. There is no explicit consensus on the approach or assumptions being made.

Contextual Notes

Some participants note that the original problem statement does not mention sigma-finiteness, raising questions about its potential impact on the solution.

onthetopo
Messages
35
Reaction score
0

Homework Statement


(X,S,u) a measure space and f is in L1.
Show that for any e>0, there exists a set E with u(E)<+infinity such that
[tex]| \int_{E} fdu - \int_{X} fdu |<e[/tex]


The Attempt at a Solution


we can define a function
[tex]v(A)=\int_{A}fdu[/tex]
It is a well known result that v(A) is in fact a signed measure.

We can somehow use the property of signed measures to show that there always exist a E such that |v(E)-v(X)|<e?
 
Physics news on Phys.org
Don't you need sigma-finiteness here or something of the sort?
 
The question is as is, there is no mention of sigma finiteness. My attempt at the solution could be completely wrong.
 
Let's suppose it is sigma finite? then what?
 

Similar threads

Replies
6
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
10
Views
3K
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K