= Solution
Let $J:C([0,1])\to L^1(0,1)$ send a <continuous function> to its <almost everywhere> equivalence class. The domain with the <supremum norm> and the target with the <L1 norm> are <Banach spaces>, and
$$
\|Jf\|_{L^1}=\int_0^1|f(t)|\,dt\leq\|f\|_\infty.
$$
Thus $J$ is a <bounded linear operator>, and the requested <vector subspace> is $J(C([0,1]))$.
To show this <image of a linear map> is not closed, consider the <continuous functions>
$$
f_n(t)=\min\{1,\max\{0,n(t-1/2)\}\}\qquad(n\geq2)
$$
and $g=\mathbf1_{(1/2,1)}$. Their <L1 norm> difference is the area of one triangle:
$$
\|Jf_n-g\|_{L^1}=\int_{1/2}^{1/2+1/n}\bigl(1-n(t-1/2)\bigr)\,dt=\frac1{2n}\longrightarrow0.
$$
But $g$ has no <continuous function> representative. Indeed, a <continuous function> equal to zero <almost everywhere> on $(0,1/2)$ must be identically zero there: any nonzero value would persist on an open interval of positive <Lebesgue measure>. Likewise any continuous representative would be identically one on $(1/2,1)$. Continuity at $1/2$ makes these requirements incompatible. Therefore $g\notin J(C([0,1]))$, although it lies in its <closure>.
The preceding <operator range category dichotomy> proves <continuous functions are meagre in L1>:
$$
\boxed{J(C([0,1]))\text{ is of the first category in }L^1(0,1).}
$$
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