Prove that $L^{\infty} [0,1]$ is not separable.
I know that $\ell^{\infty} = L^{\infty} (\Bbb N, \mathcal P (\Bbb N), \mu_c)$ is not separable where $\mu_c$ is the counting measure on $\mathcal P (\Bbb N).$ The way I proved that is to construct sequences in $\ell^{\infty}$ whose range set is contained in $\{0,1\}.$ Then such collection is uncountable and any two sequences are $1$ unit apart with respect to $\ell^{\infty}$-norm and then if we consider open balls of radius $\frac {1} {2}$ around each of the sequences then they are all disjoint and hence any dense subset of $\ell^{\infty}$ has to contain at least one element from each of these disjoint open balls. Since the number of such open balls is uncountable it follows that any dense subset of $\ell^{\infty}$ contains uncountably many elements and hence $\ell^{\infty}$ cannot be separable. How do I argue the same for $L^{\infty} [0,1]\ $?
I think of a collection of functions $\{f_{\alpha}\}_{\alpha \in I},$ for some indexing set $I$ such that $f_{\alpha} \left (\frac {1} {n} \right ) = 0\ \text {or}\ 1$ and $f_{\alpha} (x) = 0,$ for $x \not\in \left \{\frac {1} {n}\ \big |\ n \in \Bbb N \right \}.$ Then I think this collection will work here. Am I right?
Functions in $L^{\infty}$ are defined only up to a set of measure $0$. The value at $\frac 1 n$ is not well defined. So your argument does not work.
Define $T:\ell^{\infty} \to L^{\infty} ([0,1])$ by $T(a_n)=\sum_{n=1}^\infty a_n \chi_{(\frac 1{n+1}, \frac 1n)}$.
Check that $T$ is an isometric isomorphism. Can you finish?