Return to Answer

Example 4. Consider two measure spaces $(X,\Sigma_X,\mu_X)$ and $(Y,\Sigma_Y,\mu_Y)$ such that $L^\infty(X)$ and $L^\infty(Y)$ are W$^*$-algebras (equivalently, von Neumann algebras when represented in $L^2$). Such measure spaces are called localizable (sometimes an equivalent condition is used as the definition). Given a null-set reflecting measurable map $f : X \rightarrow Y$, we always get a $\sigma$-normal *-homomorphism $L^\infty(f) : L^\infty(Y) \rightarrow L^\infty(X)$, where $\sigma$-normal means that for every countable family of projections $(p_i)_{i \in I}$ in $L^\infty(Y)$ we have $$ L^\infty(f)\left(\bigvee_{i \in I}p_i\right) = \bigvee_{i \in I} L^\infty(f)(p_i). $$ We might wonder if it's always a normal *-homomorphism, where the analogous fact holds for disjoint families of projections of arbitrary cardinality (often authors restrict to $\sigma$-finite measures to simplify this away). Well, if $X$ admits a probability measure $\mu : \mathcal{P}(X) \rightarrow [0,1]$ vanishing on singletons, the answer is no. We just take $\nu$ to be the counting measure on $X$, whose only null-set is $\emptyset$, and then $\mathrm{id}_X : (X,\mathcal{P}(X),\mu) \rightarrow (X,\mathcal{P},\nu)$ is null-set reflecting and gives us a non-normal map, as the join of the family of of projections $(\chi_{\{x\}})_{x \in X}$ in $L^\infty(X,\nu) = \ell^\infty(X)$ is not preserved by $L^\infty(\mathrm{id}_X)$.

Example 4. Consider two measure spaces $(X,\Sigma_X,\mu_X)$ and $(Y,\Sigma_Y,\mu_Y)$ such that $L^\infty(X)$ and $L^\infty(Y)$ are W$^*$-algebras (equivalently, von Neumann algebras when represented in $L^2$). Such measure spaces are called localizable (sometimes an equivalent condition is used as the definition). Given a null-set reflecting measurable map $f : X \rightarrow Y$, we always get a $\sigma$-normal *-homomorphism $L^\infty(f) : L^\infty(Y) \rightarrow L^\infty(X)$, where $\sigma$-normal means that for every countable disjoint family of projections $(p_i)_{i \in I}$ in $L^\infty(Y)$ we have $$ L^\infty(f)\left(\bigvee_{i \in I}p_i\right) = \bigvee_{i \in I} L^\infty(f)(p_i). $$ We might wonder if it's always a normal *-homomorphism, where the analogous fact holds for disjoint families of projections of arbitrary cardinality (often authors restrict to $\sigma$-finite measures to simplify this away). Well, if $X$ admits a probability measure $\mu : \mathcal{P}(X) \rightarrow [0,1]$ vanishing on singletons, the answer is no. We just take $\nu$ to be the counting measure on $X$, whose only null-set is $\emptyset$, and then $\mathrm{id}_X : (X,\mathcal{P}(X),\mu) \rightarrow (X,\mathcal{P}(X),\nu)$ is null-set reflecting and gives us a non-normal map, as the join of the family of of projections $(\chi_{\{x\}})_{x \in X}$ in $L^\infty(X,\nu) = \ell^\infty(X)$ is not preserved by $L^\infty(\mathrm{id}_X)$.

This example is the cause of this answer existing, as my independent re-discovery of Fremlin's argument, for commutative W$^*$-algebras instead of measure algebras, that led to me learning what a measurable cardinal was at all. I originally made a counterexample under the assumption that all subsets of $\mathbb{R}$ were Lebesgue measurable, and then tried to find a suitable alteration of that assumption that would be compatible with the axiom of choice, which led to measurable cardinals.

This example is the cause of this answer existing, as my independent re-discovery of Fremlin's argument, for commutative W$^*$-algebras instead of measure algebras, led to me learning what a measurable cardinal was at all. I originally made a counterexample under the assumption that all subsets of $\mathbb{R}$ were Lebesgue measurable, and then tried to find a suitable alteration of that assumption that would be compatible with the axiom of choice, which led to measurable cardinals.

I have two example using measurable cardinals, and two using real-valued measurable cardinals.

I have two examples using measurable cardinals, and two using real-valued measurable cardinals.