I came up with this sequence as part of a question asking for a sequence of continuous Riemann integrable functions which converges to a function which is not Riemann integrable.

My idea was to use the fact that $\lim_{n\to \infty} e^{-n(x-a)^2}$ is the discontinuous function which is $0$ everywhere except for $x=a$, where it is $1$.

Let $e_n$ be any enumeration of the rationals in $[0,1]$. Then define a sequence $$f_n=\sum_{i=1}^n e^{-\pi n^4 (x-e_i)^2}$$

Then each $f_n$ is continuous and Riemann integrable with integral $\int_{-\infty}^{\infty} f_n=\frac 1n$ (the $\pi n^4$ in the exponential is there to make this integral nice and approach $0$, don't know if it's necessary at all). The question then is: Does this sequence converge to the Dirichlet function? And if it does, how would one prove it?

I've basically gotten nowhere. I know that I just need to show that is $x$ is irrational or is in $(-\infty,0) \cup (1,\infty)$ then $\lim_{n\to \infty} f_n(x) =0$ and that if $x$ is rational and in $[0,1]$ then $\lim_{n\to \infty} f_n(x) =1$. I just have no idea how to show either of these things, or even if they're true at all.

Any help at all would be appreciated


There is no sequence of continuous functions $(f_n)$ that converges to the Dirichlet function, which is everywhere discontinuous. This is a consequence of Baire's theorem. The limit of such a sequence is at worst pointwise discontinuous, i.e., continuous on some dense set.

There are double sequences $(f_{mn})$ of continuous functions such that

$$\lim_{m \to \infty} \lim_{n \to \infty} f_{mn}(x) = \phi(x),$$

where $\phi$ is the Dirichlet function. An example is $f_{mn}(x) = (\cos m!\pi x)^n.$

  • $\begingroup$ Wow, did not expect Baire's theorem to come up in this question, it really does pop up everywhere. Thanks a bunch for the quick and interesting response $\endgroup$ – Tsein32 Feb 5 '17 at 11:36
  • $\begingroup$ You can of course find a sequence of integrable functions that converged to the Dirichlet function, just not continuous. Let $f_n(x)=1$ on a finite subset of length $n$ of the enumeration of rationals and $0$ elsewhere. $\endgroup$ – RRL Feb 5 '17 at 11:48

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.