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I'm looking for an example of a sequence of Riemann integrable functions $(f_n)$ such that $\int_{0}^{1} f_n \rightarrow 0$ but $(f_n)$ converges to $0$ nowhere on $[0,1]$. Also, we want $f_n(x)\geq 0$ for all $x$ and $n$.

At best, I have an intuitive grasp on how to approach it. I think we need to define a sequence of subintervals $(I_n)$ of $[0,1]$ so that the length of the subinterval approaches $0$ as $n\rightarrow \infty$. Then we might be able to take the sequence of characteristic functions $(\chi_{I_n})$. Then the sequence of integrals should converge to $0$, but I'm not sure if $(\chi_{I_n})$ converges to $0$.

Is this the right idea? And if so, what is a formal proof that it works?

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    $\begingroup$ @mickep: He probably wants non-negative functions... $\endgroup$ – Beni Bogosel Feb 15 '17 at 9:43
  • $\begingroup$ @mickep Yes, nonnegative $\endgroup$ – CuriousKid7 Feb 15 '17 at 9:43
  • $\begingroup$ Try using intervals $I_{k,n} := [\frac{k-1}{n},\frac{k}{n}]$ for $k = 1, ... , n$, for each $n = 1, 2, ...$ $\endgroup$ – guest Feb 15 '17 at 9:59
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You could define the following sequence of partitions of $[0,1]$.

$P_n$: partitions into $n$ equal intervals repeated $n$ times

Then for each $n$ define $f_{n,k}$ such that $f_{n,k}$ is one on the $k$-th interval in the partition $P_n$ and zero elsewhere.

Thus the sequence of integrals is $1/n$ with eventual repetitions, but it goes to $0$. On the other hand, every $x$ is contained in infinitely many intervals on which $f_{n,k}$ is one so it $f_{n,k}(x)$ does not converge to $0$ anywhere in $[0,1]$.

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  • $\begingroup$ Could you please explain "with eventual repetitions" and why every $x$ is contained in infinitely many such intervals? $\endgroup$ – CuriousKid7 Feb 15 '17 at 11:39
  • $\begingroup$ Just construct the first examples by hand... If you partition [0,1] into n equal intervals one of these intervals contains x. Therefore given x, for any n, there exists an interval of the n-partition which contains x. Eventual repetition just means that we repeat 1/n some n times. $\endgroup$ – Beni Bogosel Feb 15 '17 at 15:15
  • $\begingroup$ How does that imply $f_{n,k}(x)$ does not converge to $0$? $\endgroup$ – CuriousKid7 Feb 15 '17 at 18:34
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    $\begingroup$ Because for each $n$, $f_{n,k}(x)$ is one for one precise $k$ (corresponding to the interval containing $x$). So the sequence $f_{n,k}(x)$ contains infinitely many zeros and infinitely many ones. $\endgroup$ – Beni Bogosel Feb 15 '17 at 23:02

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