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Can I interchange the limit and integral of a sequence of functions which is not uniformly convergent in $[0,1]$ i.e $f_n \not\to f$ uniformly is it true that $\int_0^{x_n}f_n \to \int_0^1f$ for $x_n\to 1$?

My guess is no, just thinking in terms of a picture. But if we consider $f_n(x)=x^n$ on $[0,1]$. Each function $f_n(x)$ is continuous, but the limit function $f(x)$ is not continuous: $$ f(x)=\left\{ \begin{array}{ll} 0, 0\leq x<1\\ 1, x=1\\ \end{array} \right. $$. Hence it is not uniformly convergent. But here $\int_0^{x_n}f \to \int_0^1f$ for $x_n\to 1$

Can anyone help me with one counterexample or prove it if the statement is true?

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  • $\begingroup$ What does it mean for a function to be "not uniformly convergent in $[0,1]$"? You seemed to try to clarify it by saying it's not the uniform limit of some functions, but this by itself is contradictory, as every function is the uniform limit of a constant sequence. $\endgroup$ – Brian Moehring Aug 14 at 19:04
  • $\begingroup$ Yes, you are right. I forgot to write a sequence of functions en.wikipedia.org/wiki/Uniform_convergence. My apology! $\endgroup$ – Shadow Aug 14 at 19:23
  • $\begingroup$ What picture are you thinking in terms of? $\endgroup$ – Aaron Aug 14 at 19:38
  • $\begingroup$ Uniform convergence is sufficient but not necessary (which you just demonstrated with a counterexample). $\endgroup$ – RRL Aug 14 at 19:44
  • $\begingroup$ That edit doesn't really resolve the issue. In order for the sequence $f_n$ to matter at all you need to either put some restrictions on the type of functions given by $f_n$ or you need to include the $f_n$ in the integral, as in perhaps $$\int_0^{x_n} f_n \to \int_0^1 f$$ Otherwise, we might as well not mention a sequence at all, and the necessary and sufficient condition is simply that $f$ integrable on $[0,1]$ $\endgroup$ – Brian Moehring Aug 14 at 19:50
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If you have a sequence of continuous functions that don't converge uniformly, then they might not converge to a continuous function, and depending on what definition of integration you are using, that might be a problem. However, limits of measurable functions are measurable, so if you are doing measure theory, things are okay.

Limits commute with integrals in nice cases. For example, there is the monotone convergence theorem and the dominated convergence theorem, and there is a discussion of why the conditions are actually necessary on those pages. But you may want to take those theorems as hints: since they are true, they tell you something about what a sequence of functions must look like to produce a counterexample.

But I will give the following partial hint to keep you from looking at those pages: You can find an example of continuous functions that converge pointwise to 0 on $[0,1]$ but every integral is equal to $1$.

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  • $\begingroup$ Hi, I have calculated this one Let $E_{k}=(1-1/2^{k},1-1/2^{k+1}]$, let $f_{n}$ be the functions defined in $[0,1]$ by $$ f_{n}(x) = \sum_{k=n}^{\infty} 2^{k+1}\chi_{E_{k}}(x), $$ where $\chi_{E_{k}}$ is the characteristic function of $E_{k}$. I will choose $x_n\in(1-1/2^{k+1},1)$. The sequence $\{f_{n}\}$ decreases pointwise to the zero function, but the integrals are all infinite. $\endgroup$ – Shadow Aug 14 at 20:56
  • $\begingroup$ You can give me any other answer if you have in mind. I am thinking that you are trying to play with $$\sum \frac 1{2^n}$$, not sure though!! $\endgroup$ – Shadow Aug 14 at 20:57
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    $\begingroup$ @Shadow Even simpler, $f_n=2^{n+1}\chi_{E_n}$.No summing necessary. $\endgroup$ – Aaron Aug 14 at 22:14

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