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?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ 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 $\infty$. $\endgroup$– Rhys SteeleOct 10, 2019 at 15:14
-
$\begingroup$ Yes, you are right. I was thinking of a $\sigma$-additive finite measure. $\endgroup$– Condor5Oct 10, 2019 at 16:09
-
$\begingroup$ In general, editing a question to add conditions that invalidate an answer already posted is bad practice on this site. $\endgroup$– Rhys SteeleOct 10, 2019 at 22:17
-
$\begingroup$ Sorry. I edited the question. $\endgroup$– Condor5Oct 10, 2019 at 23:14
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
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 $\mu(T) = \infty$.
answered Oct 10, 2019 at 16:30
-
$\begingroup$ This works. Is there an example in which $T$ is a compact space? $\endgroup$– Condor5Oct 10, 2019 at 17:14
-
$\begingroup$ I am precisely thinking of cases in which the family of measures that you propose is not continuous with respect to the total variation norm. $\endgroup$– Condor5Oct 10, 2019 at 17:22
-
$\begingroup$ I don't have time to think more about this right now and don't have an example straight away. In future, it'd probably be helpful for you understanding these things if you try to think about these edge cases and work out what conditions you want to impose and include them in the question so that you don't get answers like mine. $\endgroup$ Oct 10, 2019 at 22:15
-
$\begingroup$ Your answer is very helpful. I voted it up. I am wondering if there is also an answer with a compact space. I’ll edit the question to clarify. $\endgroup$– Condor5Oct 10, 2019 at 23:11