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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

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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 $f_{mn}(x) = (\cos m!\pi x)^n.$

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  • $\begingroup$ 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

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