Commutant of $C_0(X)$ inside $\ell^2(X)$. Let $X$ be a locally compact Hausdorff space.
Consider the commutative $C^*$-algebra $C_0(X)$ and implement it as a $C^*$-subalgebra of $B(\ell^2(X))$ via multiplication operators:
$$L: C_0(C)\to B(\ell^2(X)): f \mapsto (g\mapsto fg)$$
What is the commutant (and by extension the bicommutant) of the image of this representation? I conjectured it is
$$C_0(X)'=\ell^\infty(X).$$
Clearly the right hand side is contained in the left hand side, but conversely I should show that if $t$ commutes with $C_0(X)$, then $t$ is multiplication with some bounded function. Constructing this bounded function is what gives me trouble.

I don't know how to find the commutant, but I know it is usually strictly bigger than $\ell^\infty(X)$.
For instance let $X=[0,1]$, with the usual topology. Let $E\subset X$ be a non-Lebesgue-measurable set. Let $h=1_E\in\ell^\infty(X)$. If $h\in C_0(X)''$, then by the Double Commutant Theorem there exists a net $\{f_j\}\subset C_0(X)$ such that $f_j\to h$ in the sot. This means that $f_jg\to hg$ in $\ell^2(X)$ for all $g$. That is,
$$
\lim_j\sum_{x\in X} |f_j(x)-h(x)|^2\,|g(x)|^2=0,\qquad g\in\ell^2(X). 
$$
This in particular implies that $f_j\to h$ pointwise. But a pointwise limit of continuous functions is measurable, a contradiction.
So $C_0(X)''\subsetneq\ell^\infty(X)$, which implies $\ell^\infty(X)\subsetneq C_0(X)'$.
As $\ell^\infty(X)$ is maximal abelian, this means that $C_0(X)'$ is non-abelian.
