space of riemann integrable functions not complete

Solution 1:

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.