# Total variation versus absolute continuity

Can you given an example of a family of measures $$\{\mu_t\}$$, with $$t\in T$$, $$T$$ a topological space, that are continuous in $$t$$ with respect to the total variation norm but not absolutely continuous with respect to any $$\sigma$$-additive finite measure? Is there an example when $$T$$ is compact or Lindelof?

• You're probably going to want to add some conditions on this statement. Right now, the answer is trivially no since every measure on a given measurable space is absolutely continuous with respect to the measure that gives every non-empty set measure $\infty$. Oct 10, 2019 at 15:14
• Yes, you are right. I was thinking of a $\sigma$-additive finite measure. Oct 10, 2019 at 16:09
• In general, editing a question to add conditions that invalidate an answer already posted is bad practice on this site. Oct 10, 2019 at 22:17
• Sorry. I edited the question. Oct 10, 2019 at 23:14

Let $$T$$ be an uncountable space equipped with the discrete topology. In particular, for any topological space $$Y$$, every function $$f:T \to Y$$ is continuous.
Hence, we are free to choose any family of measures $$\{\mu_t\}$$ on $$(T, B(T))$$ and $$t \mapsto \mu_t$$ will be continuous for the total variation norm. In particular, we can take $$\mu_t = \delta_t$$ so that $$\mu_t(A) = 1_{t \in A}$$.
Now suppose that $$\mu$$ is a measure such that $$\mu_t$$ is absolutely continuous with respect to $$\mu$$ for every $$t$$ then $$\mu$$ must assign positive measure to every non-empty Borel subset of $$T$$. Since $$T$$ has the discrete topology, this means that $$\mu(\{t\}) > 0$$ for every $$t \in T$$ and hence that $$\mu(T) = \infty$$.
• This works. Is there an example in which $T$ is a compact space? Oct 10, 2019 at 17:14