L^1 convergence and uniform convergence

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SUMMARY

The discussion focuses on the concepts of L^1 convergence and uniform convergence within the context of measure theory, specifically using the Lebesgue measure on the real line. An example is provided where a sequence of functions {f_n} converges uniformly to f = 0, yet does not converge in L^1 norm, illustrating that uniform convergence does not imply convergence in L^1. The L^1 norm is defined as \| f\|_{L^1(X,\mu)}=\int_X |f|d\mu, emphasizing the distinction between these types of convergence.

PREREQUISITES
  • Understanding of measure spaces, specifically Borel measure.
  • Familiarity with Lebesgue measure and its properties.
  • Knowledge of convergence concepts in functional analysis, particularly uniform and L^1 convergence.
  • Basic understanding of integrable functions and the definition of the L^1 norm.
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  • Study examples of sequences of functions that converge uniformly but not in L^1.
  • Explore the implications of the Dominated Convergence Theorem on convergence types.
  • Learn about other L^p spaces and their convergence properties.
  • Investigate the relationship between pointwise convergence and uniform convergence in measure theory.
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Mathematicians, particularly those specializing in analysis, graduate students studying measure theory, and anyone interested in the nuances of convergence in functional spaces.

quasar987
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Let (X,µ) be a measure space with µ(X) infinite. I'm trying to find an example of a pair (X,µ) and a sequence {f_n} of L^1(X,µ) functions converging uniformly to a function f such that we do not have f_n --> f in L^1(X,µ).
 
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If your measure space is Borel then I don't think this statement is true. A sequence of Borel measurable functions that has a limit converges to a Borel function.
 
RedX said:
If your measure space is Borel then I don't think this statement is true. A sequence of Borel measurable functions that has a limit converges to a Borel function.
I don't see how you can conclude the statement is false from this.

In fact it is possible to come up with an example satisfying the OP using X=R with Lebesgue measure. Take f_n to be a sequence of rectangles with decreasing height but increasing width, in such a way that f_n -> 0 uniformly but not in L^1. You can even rig it so that f_n doesn't converge in any L^p (1 <= p < inf).
 
morphism said:
In fact it is possible to come up with an example satisfying the OP using X=R with Lebesgue measure. Take f_n to be a sequence of rectangles with decreasing height but increasing width, in such a way that f_n -> 0 uniformly but not in L^1. You can even rig it so that f_n doesn't converge in any L^p (1 <= p < inf).

Doesn't L^1 mean the space of integrable functions? So why isn't f=0 in this space?
 
RedX said:
Doesn't L^1 mean the space of integrable functions? So why isn't f=0 in this space?

Yes, f(x) = 0 is in L^1. The issue is whether f_n converges to f in the L^1 norm.
 
Ah! Thank you morphism.
 
kilimanjaro said:
Yes, f(x) = 0 is in L^1. The issue is whether f_n converges to f in the L^1 norm.

What is the L^1 norm?
 
The L^1 norm is defined on the space of all integrable functions as
\| f\|_{L^1(X,\mu)}=\int_X |f|d\mu<br />
 

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