Let $\{r_1,r_2,…,r_n,…\}$ be an enumeration of the rational numbers in $I:=[0,1]$ and define $f_n:I \to \mathbb{R}$ by $f(x) =1 , $ if $x = r_1,…,r_n$ and $f(x) = 0,$ otherwise. Find $\lim(f_n(x)),$ if it exists.
May I verify if my proof is correct?
Proof: Since $|\mathbb{Q}|=|\mathbb{N}|, \exists n \in \mathbb{N} $ such that $f_n(1)=1. $ Also, observe that $f_i(x) \leq f_j(x) \leq 1,$ whenever $i \leq j$ and $\forall x \in I.$
This means $\lim_{n \to \infty} f_n = f:I \to \mathbb{R},$ where $ f(x) = 1 \ (x \in \mathbb{Q} \cap I) $ or $ 0 $ (otherwise).
Lastly, since $f_n$ has finite number, $n$ of discontinuities, then $f_n \in \mathscr{R}[0,1]?$
Thank you.
Best Answer
You've more or less proven pointwise convergence of $f_n(x)$ to $f(x)$ for all x in $[0,1]$. On the other hand the sequence of functions is not uniformly convergent: for every integer $N$ there are an infinite number of points $x = r_n: n>N$ for which $f_N(x) - f(x) = 1$, and so the rate of convergence of $f_n(x)$ to $f(x)$ depends on the argument $x$.
As for Riemann integrability, $f_n$ is discontinuous only at $r_1, \ldots,r_n$, and so is continuous almost everywhere. As it is also bounded, it is Riemann integrable. $f$ however is nowhere continuous, so not Riemann integrable.