Extension of Lebesgue Convergence Theorem

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SUMMARY

The discussion focuses on the extension of the Lebesgue Convergence Theorem, specifically addressing problem 4.14 from Royden's text. The key conclusion is that if a sequence of integrable functions converges to an integrable function f almost everywhere, then the integral of the absolute difference, \int |fn - f|, converges to 0 if and only if \int |fn| converges to \int |f|. The theorem states that if the sequence bounds and converges to g, then the integral of f equals the limit of the integrals of .

PREREQUISITES
  • Understanding of Lebesgue integration
  • Familiarity with the concept of almost everywhere convergence
  • Knowledge of integrable functions and their properties
  • Proficiency in applying inequalities in analysis, specifically the triangle inequality
NEXT STEPS
  • Study the proof of the Lebesgue Dominated Convergence Theorem
  • Explore the implications of the Dominated Convergence Theorem in real analysis
  • Learn about the properties of integrable functions in the context of measure theory
  • Investigate examples of sequences of functions that converge almost everywhere
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Mathematicians, graduate students in analysis, and anyone studying measure theory or functional analysis will benefit from this discussion.

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Homework Statement


This comes courtesy of Royden, problem 4.14.
a.Show that under the hypothesis of theorem 17 we have \int |fn-f| \rightarrow 0

b.Let <fn> be a sequence of integrable functions such that fn \rightarrow f a.e. with f integrable. Then \int |fn-f| \rightarrow 0 if and only if \int |fn| \rightarrow \int |f|

Homework Equations


Theorem 17:
Leg <gn> be a sequence of integrable functions which converges to an integrable function g. Let <fn> be a sequence of measurable functions such that |fn|\leq gn and <fn> converges to f a.e. If
\int g = lim \int gn
Then
\int f = lim \int fn



The Attempt at a Solution


My problem in part a is the same as the "if" in part b. The only if seems fairly trivial since |a-b|>=| |a| - |b| |. Unfortunately that is partially the problem. Every attempt I've made to go in the other direction reduces to that same inequality. My basic thought process far has been to start with
\int |f| = lim \int |fn| which we have by assumption, and then conclude that for some e>0 there's an N such that
| \int |fN| - |f| |&lt; e

Obviously this does not get me where I want to go. If I somehow had fn as an increasing function I could keep going, but I see no real way to do this.
 
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After poking at it a bit more, if I could establish the same inequality that I'm arriving at on my last step for the portions of f and fn where f>=fn and where f <=fn then I could get it from there. (that is, split |f|-|fn| into positive and negative portions.)
 
Figured it out. If we define Fn to be f if f<fn and fn if fn<f, then we have that |Fn|<|f|+|fn|, which we is integrable and the limit of it's integral is 2 times the limit of the integral of |f|.
Since Fn converges a.e. to f, we then have that the integral of f is equal to the limit of the integral of Fn. Then we very quickly get that the integral of f-Fn is equal to the integral of |f-Fn| (since Fn is less than or equal to f, as well as fn) and for any n>N this is less than e. Then we also have that the limit of the integral of fn and Fn are the same, so the integral of fn-Fn converges, and is positive for every n, so the integral of |fn-Fn| is equal to it and less than e for n>m for some m. Then lastly we have that the integral of |f-fn|=integral of |f-Fn +Fn -fn| which is less than or equal to the integral of |f-Fn| + integral of |fn-Fn| which are less than e for each, so the whole thing is less than 2e, so it converges.

Sorry if that wasn't clear but my brain is now fried and I'm going to bed.:smile::zzz:
 

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