Can a Measure be Literally Continuous?

Question

Say that $(X,M)$ is a measurable space (i.e., $0,X \in M$ and $A \in M \implies X \setminus A \in M$ and, finally, $M$ is closed w.r.t. countable unions/intersections). Suppose that $\mu$ is either a signed or complex measure on $(X,M)$. In particular, $\mu$ is a function on $M$.

I know that if $\mu$ maps into $[0,\infty)$, then $d(A,B) = \mu(A \Delta B)$ is a metric modulo the equivalence relation $A \sim B \iff \mu(A\Delta B) = 0$. This was mentioned in baby Rudin, for instance.

More generally, however, if we are given a topology $T$ on $X$ does there consequently exist a canonical topology on $M$ induced by $T$ such that $\mu$ is continuous?

Of course, I would prefer if the topology on $M$ were not just the discrete topology.

Further, it would be nice if it were metrizable. If the answer is affirmative, a follow up question might be, for instance, "how coarse can the topology be while still making $\mu$ continuous?" Has anyone thought about this or seen this before?

Thanks.

Answer 1

You could also take the pseudometric in $M$ such that $d(A,B)=|\mu(A)-\mu(B)|$, this is obviously continuous since $\mu^{-1}(B_{\mathbb C}(\mu(A),\varepsilon))=B_{M}(A,\varepsilon)$ and this is the weakest topology in $M$ such that $\mu$ is continuous, since $\tau_d=\{\mu^{-1}(A):A\in \tau_{\mathbb C}\}$, one property this topology has is that sets $A,B$ such that $\mu(A)=\mu(B)$ are not separable.

Answer 2

Take the pseudometric $d$ on $M$ such that $d(A,B)=|\mu|(A\Delta B)$. Take $B(a,r)\subset \mathbb C$ and a set $A\in M$ such that $\mu(A)\in B(a,r)$, also take $\varepsilon>0$ such that $B(\mu(A),\varepsilon)\subset B(a,r)$. Take $B$ in the ball of radius $\varepsilon$ centered in $A$, so $|\mu|(A\Delta B)=|\mu|(A\setminus B)+|\mu|(B\setminus A)<\varepsilon$. Now $$|\mu(A)-\mu(B)|=|\mu(A\setminus B)-\mu(B\setminus A)|\leq |\mu(A\setminus B)|+|\mu(B\setminus A)|\leq|\mu|(A\setminus B)+|\mu|(B\setminus A)=|\mu|(A\Delta B)<\varepsilon$$ So $\mu(B(A,\varepsilon))\subset B(\mu(A),\varepsilon)\subset B(a,r)$. This implies that every point in the preimage of $B(a,r)$ is an interior point, so $\mu^{-1}(B(a,r))$ is open which implies that $\mu:(M,d)\to(\mathbb C,\tau_{\mathbb C})$ is continuous.