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?

  • 1
    $\begingroup$ @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

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 $[0,1]$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.