Finding a maximal complete subspace of Riemann Integrable functions on $[0,1]$ I know that the space of Riemann Integrable functions on $[0,1]$ is not complete under the norm $|f|= \int f$. So I was wondering as to what would be a maximal complete subspace of Riemann Integrable functions. 
My Try: Firstly I noticed that the subspace of constant functions is complete. So we know that the set of all complete subspaces is not empty. Now I would like to show that any chain of complete subspaces has a maximal element (I am having trouble showing this also I am not sure if this is true). 
If we succeed then by Zorn's lemma there is definitely a maximal subspace. So we would have shown its existence.
Edit: I know that Lebesgue Integrable functions are in a way a completion of Riemann Integrable functions. So I was probably thinking we could take some subspace which is contained in Riemann Integrable functions then we would get some other spaces.
But I am wondering if we can actually write down the set.Any hints or comments are welcome. Thanks.
 A: There is no maximal complete subspace in the space of Riemann-integrable functions.
In fact, given any non-complete normed space $X$, there is no maximal complete subspace in $X$.
To see this, asumme towards a contradiction that such a subspace exists; call it $E$. Then $E\neq X$ since $X$ is not complete. Take any $u\in X\setminus E$, and consider $F:= E\oplus [u]$, where $[u]$ is the $1$-dimensional subspace spanned by $u$. Let us show that $F$ is complete, which will contradict the maximality of $E$.
Consider the projection map $p:F\to [u]$ associated with the decomposition $F=E\oplus [u]$. The range of $p$ is $1$-dimensional, and the kernel of $p$ is $E$ which is a closed subspace of $F$ since $E$ is assumed to be complete. It follows that $p$ is continuous (using the well known fact that a linear functional is continuous if and only if its kernel is closed). So $Id-p$ is continuous as well. Hence, we have estimates $\Vert p(x)\Vert\leq C\Vert x\Vert$ and $\Vert x-p(x)\Vert\leq C\Vert x\Vert$ for $x\in F$. Since both $E$ and the $1$-dimensional space $[u]$ are complete, it now follows easily that $F$ is complete: if $(x_n)$ is a Cauchy sequence in $F$, then $(p(x_n))$ and $(x_n-p(x_n))$ are Cauchy in $[u]$ and $E$ respectively, etc.
