Lim fn(x) = f(x) but lim ∫ |f(x)-fn(x)|dx ≠ 0 ?

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Homework Help Overview

The discussion revolves around finding a sequence of integrable functions \( f_n: \mathbb{R} \to \mathbb{R} \) that converges pointwise to a function \( f \), while the integral of the absolute difference does not converge to zero. The original poster seeks examples that satisfy these conditions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the function \( f_n = x + \frac{x}{n} \) and questions the interchangeability of limits and integrals. Another participant suggests a different function \( f_n(x) = |1/n| \) for \( x \in [0, n] \) and \( f_n(x) = 0 \) elsewhere, noting its uniform convergence to 0.

Discussion Status

Participants are exploring various function examples and discussing the conditions under which limits and integrals can be interchanged. Some guidance has been offered regarding the validity of the proposed functions, but there is no explicit consensus on a definitive example yet.

Contextual Notes

The original poster seeks clarification on the term 'integrable' in the context of Riemann integrability and is considering the implications of boundedness in their attempts.

pulin816
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Homework Statement



Hey, I have another questions,
I need to find an example of a sequence of integrable functions fn:R -> R, n =1, 2, ...
such that

lim fn(x) = f(x) (as n -> ∞)
but lim ∫ |f(x)-fn(x)|dx ≠ 0 (as n -> ∞)​
(with integral from - to + infinity)

The Attempt at a Solution



I've tried
fn = (x + x/n)
and f = x

the first conditions would be satisfied, but on the other hand,
will the limits and the integral be interchangeable? I've read that it is only permitted if the expression inside is bounded. |x/n| can't be bounded since it has an absolute sign wrapped around or would it?

Any suggestions? Thank you !

p.s. Would the term 'integrable' here mean a function that is reinmann integrable?
 
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I wouldn't worry about the interchange of limits, at least not right now. The function you suggested would not work because the limit of the integrals is 0 anyway.
 
Thanx LeonhardEuler:

I tried coming up with a nother function, what about

fn(x) = |1/n| for x ∈ [0, n] and
fn(x) = 0 elsewhere in [-infinity, infinity]​
fn(x) will converge to 0 function uniformly
however, while ∫ |fn(x)| dx = 2 for all n in N

?
 
Yeah, that works.
 
Except the integral is 1, not 2.
 
oh yeah, my mistake.
Thanks!
 

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