Let $\{\Gamma_i , \mu_i\}_{i\in I}$ be a family of probability measure spaces and suppose $I$ is uncountable. Let $\{\Gamma , \mu\} = \prod_{i\in I} \{\Gamma_i,\mu_i\}$ be the product measure space. Set $H=L^2(\Gamma , \mu)$. Let $M$ be the von Neumann algebra on $H$ generated by multiplication operators $m_\phi$ by bounded measurable functions $\phi$. Show that the constant function $1$ is a separating and cyclic vector for $M$, yet $H$ is not separable.
For bounded measurable function $\phi$, $m_\phi(1) = 0$ implies that $\phi= 0$, so constant function $1$ is separating. But I can not show $\overline {M(1)} = H$. Please help me. Thanks in advance.
The assertion $\overline{M1}=H$ is the assertion $\overline{L^\infty(\Gamma,\mu)}=L^2(\Gamma,\mu)$, i.e. that every $L^2$ function is an $L^2$-limit of $L^\infty$ functions. This is a canonical exercise in measure theory: given $g\in L^2$, let $E_n=\{|g|^2\geq n\}$. Then $\mu(E_n)\to0$, because $$ \mu(E_n)=\int_{E_n}1\,d\mu=\frac1n\,\int_{E_n}n\,d\mu\leq\frac1n\,\int_{E_n}|g|^2\,d\mu\leq\frac{\|g\|_2^2}n. $$ Note that $E_{n+1}\subset E_n$, so the sequence $\{1_{E_n}\}$ is monotone; as $\mu(E_n)\to0$, we deduce that $1_{E_n}\to0$ pointwise $a.e.$
Now let $g_n=g\,1_{E_n^c}\in L^\infty$; then, using dominated convergence, $$ \|g_n-g\|_2^2=\int|g_n-g|^2\,d\mu=\int 1_{E_n}|g|^2\,d\mu\to0. $$