I am looking for a normed space whose completion has strictly larger cardinality. I have settled on the space $I^1[0,1]$ of simple functions on $[0,1]$ with completion $L^1[0,1]$ the space of equivalence classes of measurable functions. The fact that they are equivalence classes is important.
I believe $I^1[0,1]$ has cardinality $\frak c$ since we can represent each simple function as an increasing finite sequence corresponding to the discontinuity points, and a second sequence corresponding to the output values. So we get at most $|\mathbb R|^{|\mathbb N|} \times |\mathbb R|^{|\mathbb N|} = |\mathbb R| = \frak c$ elements.
I am convinced that there are more Lebesgue functions than there are indicator functions but cannot see why. One thing that will not work is taking indicator functions of Borel sets. I am pretty sure there are only $\frak c$ Borel sets, since they are generated by intervals with rational endpoints. Performing the countable-input sigma operations we can only get $2^{|\mathbb N|} $ different elements.
Something else that won't work is choosing a Lebesgue set of measure zero and considering all subsets. These will certainly be measurable sets, and there are as many of them as we need. But the corresponding indicator functions are all equivalent outside a set of measure zero. So we only get a single element of $L^1[0,1]$
Does anyone know the cardinality of $L^1[0,1]$?
It is $\frak c$. Hint: Each $f$ in $L^{1}[0,1]$ is the limit of a sequence of polynomials with rational coefficients.