# Can a Measure be Literally Continuous?

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.

• The weaker topology you can get such that $\mu:M\to \mathbb C$ is continuous is the topology on $M$ generated by $\lbrace \mu^{-1}(B(z,r)):z\in\mathbb C, r>0\rbrace$. Dec 17, 2020 at 5:37

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.
• Two great answers! I will need to check that the induced topology $\tau_d$ indeed satisfies $\tau_d = \{\mu^{-1}(A) : A \in \tau_{\mathbb{C}}\}$, as you say. Without yet checking this, however, that seems reasonable. (so many edits to this comment because I can't be bothered to proof read the Mathjax :) ) Dec 17, 2020 at 17:51
• since $\tau_\mathbb C$ is generated by the open balls and $f^{-1}(\bigcup A_i)=\bigcup f^{-1}A_i$ you only need to check that for the open balls wich is given by definition. Dec 18, 2020 at 16:38
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.