Noncomplete example of Riemann-Integrable function spaces I'm finding some noncomplete examples of Riemann-Integrable function spaces $\mathcal R^1[0,1]$ and $\mathcal R^2[0,1]$, when the given norms are $\|f\|_1=\int_0^1 |f|$ and $\|f\|_2=(\int_0^1|f|^2)^{1/2}$.
My solution was :
We can construct a sequence $\{q_n\}$ in $\mathbb Q$, which consists of distinct rationals. Consider a Cauchy sequence $f_n=\chi _{\{q_1,q_2,...q_n\}}\in \mathcal R^1[0,1]\cap\mathcal R^2[0,1]$. Since $\mathbb Q$ is countable, we can consider $f=\chi _{\{q_1,q_2,q_3,...\}}=\chi_{\mathbb Q}$ and $f_n$ should converge to $f$, but $f$ and $f^2$ are not Riemann-integrable so not in both spaces. So both spaces are not complete.
I think considering $\{q_n\}$ and limiting $n$ to infinity is not very clear, but cannot get over it. Can I ask some help for this? Any other good examples will be appreciated.
 A: In the sense of metric space, if $\|f - g \|_1=\int_0^1 |f-g| = 0$, then $f$ and $g$ should be the same point in the space (definition of metric).
To take $\mathcal R^1[0,1]$ as metric space one should identify such functions $f$ and $g$, considering an equivalence class of functions where $\|f-g\|_1=0$ $\implies$ $[f]=[g]$ rather than the usual function equality.
Pointwise convergence of a sequence of functions is not the same as convergence in the metric space $\mathcal R^1[0,1]$. Using the $\epsilon$ definition of limits of sequences gives the following:
Let $g(x)=0$ for all $x$, then for all $\epsilon>0$
$$\|f_n-g\|_1 = \int_0^1 |f_n-g| = 0 < \epsilon.$$
In fact, the sequence $f_n$ in the metric space $\mathcal R^1[0,1]$ is a constant sequence (all functions in the sequence are equivalent) that converges to the class of functions in $\mathcal R^1[0,1]$ that is almost always zero.
A better approach to prove the incompleteness of $\mathcal R^1[0,1]$ is think about unbounded functions. Recall that Riemann-integrable means bounded and that the set of discontinuities is a null set. A good start is the following sequence of functions defined in the closed interval $[0,1]$:
$$f_n(x) = \min \left(n, \frac{1}{\sqrt{x}}\right)$$
