L^1 convergence and uniform convergence

[quasar987]

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).

 

[RedX]

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.

 

[morphism]

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).

 

[RedX]

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?

 

[kilimanjaro]

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.

 

[quasar987]

Ah! Thank you morphism.

 

[RedX]

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?

 

[quasar987]

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