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Is it possible to find:

Sequence of continuous functions $f_n:[0,1]\rightarrow \mathbb{R}$ that converges to the zero function and such that sequence $\int_0^1 f_1(x)dx, \int_0^1 f_2(x)dx,\ldots$ increases without a bound

I think it's quite easy. Just define $f_n$ in the following way:

enter image description here

Uniformly convergent sequence of differentiable functions $f_n : (0,1)\rightarrow \mathbb{R}$ such that the sequence $f_1 ', f_2',\ldots$ does not converge.

Here I have trouble.

Convergent sequence of Riemann integrable functions $f_n : [0,1]\rightarrow \mathbb{R}$ whose limit function IS NOT Riemann integrable.

I know the example that uses charactersitic function of rationals. But is it possible to give another example?

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    $\begingroup$ Your first example doesn't converge at $0$; but that's easily fixed ( /\ ). $\endgroup$ – David Mitra Dec 14 '15 at 16:35
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    $\begingroup$ cHere I wrote up some continuous functions that have constant integral but converge pointwise to zero: math.stackexchange.com/questions/1553955/… If you multiply the function $f_n$ by n pointwise convergence to zero will still be given but the integral will diverge. $\endgroup$ – s.harp Dec 14 '15 at 16:35
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    $\begingroup$ Just make the graph look like what I wrote at the end of my first comment. $\endgroup$ – David Mitra Dec 14 '15 at 16:52
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    $\begingroup$ The triangle will have vertices at the origin, $(1/(2n),2n^2)$, and $(1/n,0)$. Its area will be ${1\over 2}\cdot 2n^2\cdot{1\over n}=n$. $\endgroup$ – David Mitra Dec 14 '15 at 17:05
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    $\begingroup$ And it will still converge to $0$, of course. $\endgroup$ – David Mitra Dec 14 '15 at 17:07
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(second question) $f_n(x)=\frac1n\sin(n^2x)$

(third question) $f_n(x)=1/x$ on $[1/n,1]$ and $n^2x$ on $[0,1/n].$

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  • $\begingroup$ (i) has finite integral on $[0,1]$, (ii) is not continuous. $\endgroup$ – s.harp Dec 14 '15 at 16:46
  • $\begingroup$ It is not clear from the question that the integral condition applies to the second question and that the continuity requirement applies to the third question. $\endgroup$ – Justpassingby Dec 14 '15 at 16:48
  • $\begingroup$ Ah, I think you messed up your numbering, (i) is (ii) and (ii) is (iii) ;) $\endgroup$ – s.harp Dec 14 '15 at 16:51
  • $\begingroup$ I still managed to make the last example continuous :-) $\endgroup$ – Justpassingby Dec 14 '15 at 16:52
  • $\begingroup$ Each question is separate :) Thanks. $\endgroup$ – luka5z Dec 14 '15 at 16:52

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