 Continuous functions are meagre in L1

The image of $C([0,1])$ in the L1 space $L^1(0,1)$ is a meagre set. The canonical map $J$ is a bounded linear operator from the Banach space $C([0,1])$ with its supremum norm, since $\Vert Jf{\Vert }_1\le \Vert f{\Vert }_{\infty }$. It is not onto: the indicator of $(1/2,1)$ has no continuous function representative. Continuity would force values zero throughout $(0,1/2)$ and one throughout $(1/2,1)$, contradicting continuity at $1/2$. The operator range category dichotomy therefore proves the claim. Continuous ramps $f_n(t)=\min (1,\max (0,n(t-1/2)))$ also show directly that this image is not closed, since their L1 norm distance to that indicator is $1/(2n)$ for $n\ge 2$.