Define norm as $\int |f|$ (Riemann integral) on $\mathcal R^1[0,1]$, the space of riemann integrable functions on $[0,1]$ with identification $f=g$ iff $\int |f-g|=0$.
Let $\{ r_1,r_2,\cdots \}$ be the rationals in $[0,1]$, and let $f_n=1_{\{r_1,\cdots,r_n\}}$. Then $f_n$ is a cauchy sequence in $\mathcal R^1[0,1]$. I want to show that there is no $f\in \mathcal R^1[0,1]$ such that $f_n$ converges to $f$ in norm. How can I show it?
Obviously the pointwise limit $f=1_\mathbb{Q}$ is not contained in $\mathcal R^1[0,1]$, but can I use this fact? I think that there can be other candidates, since convergence in norm and pointwise convergence are different.
Best Answer
Recall the condition that $f=g$ if and only if $\int|f-g| = 0$. This means that elements of $\mathcal R^1$ are not functions in the classical sense, because they're only defined up to sets of measure $0$. You can't evaluate $f(x)$, because every pair $(x,y)$ there's some $g\in\mathcal R^1$ with $g=f$ but $g(x)=y$. We just change the value of $f$ at a single point.
So in your example we have $f_n = 0$ for every $n$.
Consider instead the functions
$$g_n(x) = \min\left(n,-\log x\right)$$.
Now each $g$ is Riemann integrable, and it's easy to see that the sequence is Cauchy $\int|g_n - g_m|\leq \int|g_n| - 1$ for $m>n$.
But there is no limit in $\mathcal R^1$. If there is a limit it must be $x\mapsto-\log(x)$ (almost everywhere), but that isn't Riemann itegrable because it's unbounded.