I read a proof about this topic at this site and was convinced with it , then I tried to construct a cauchy sequence of functions to see how things apply . Now I'm confused.
The sequence I constructed is : $f_n(x)=\left\{\begin{matrix} -lnx & if & x> e^{-n}\\ n& if & x\leq e^{-n} \end{matrix}\right.$
I proved that this sequence is cauchy and that it belongs to the mentioned space . For the limit I got that $f_n(x)\rightarrow f(x)=-lnx$
almost everywhere. But $f$ is not bounded on $\left [ 0,1 \right ]$
Did I miss or mess with something ??
First observe that the space $R[0,1]$ of all Riemann integrable functions is a subspace of the complete space $B[0,1]$ formed by all bounded functions. So $R[0,1]$ is complete iff it is closed in $B[0,1]$.
Theorem. $R[0,1]$ is closed in $B[0,1]$, and hence complete.
Proof. Given a sequence $\{f_n\}_n$ in $R[0,1]$ converging to some $f$ in $B[0,1]$, let $D_n$ be the set of points where $f_n$ is discontinuous.
It is well known that $D_n$ has measure zero, and hence so does $$ D := \bigcup_n D_n. $$
Notice that if $x$ is not in $D$ then each $f_n$ is continuous at $x$, and hence so is $f$. In other words the discontinuities of $f$ lie in $D$ so $f$ is Riemann integrable. QED