I have a question in measure theory: given two measures $\nu$ and $\mu$, we say that $\nu$ is absolutely continuous of $\mu$ if for Borel set $A$ such that $\mu(A)=0$, we have $\nu(A)=0$. I want to know if it is possible that $\mu$ is a measure for a continuous random variable and $\nu$ is a measure for a discrete random variable so that $\nu$ is absolutely continuous of $\mu$, or vice versa?
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2$\begingroup$ What steps have you taken to answer your question? This ought to be a (simple) matter of applying the definition--but that will depend on what definitions of "discrete random variable" and "continuous random variable" you are working with: what are they and how have you used them? $\endgroup$– whuber ♦Commented Jun 18, 2019 at 14:00
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$\begingroup$ I want to know if it is general true or not. Indeed, I am working on f-divergence. The definition requires that one measure is absolute continuous w.r.t. another measure. But I also read some papers which state that some f-divergence can include discrete distribution in a divergence ball centered at a continuous random variable. That's why I wonder if there exist a discrete probability which is absolute continuous w.r.t. continuous probability. $\endgroup$– will_cheukCommented Jun 19, 2019 at 2:37
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1 Answer
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A continuous random variable is a random variable whose cumulative distribution function is continuous everywhere. This implies that the associated measure $\mu$ puts zero weight on any single value $\mathfrak s$, $\mu(\{\mathfrak s\})=\mu(\mathfrak s)=0$. A discrete random variable takes its values in a discrete or countable set, $\mathfrak S$, meaning that the associated measure $\nu$ puts weight one on $\mathfrak S$: $$\sum_{\mathfrak s\in\mathfrak S} \nu(\mathfrak s)=1$$ while $$\sum_{\mathfrak s\in\mathfrak S} \mu(\mathfrak s)=0$$
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2$\begingroup$ So the answer is that it is impossible, right? $\endgroup$ Commented Jun 18, 2019 at 6:14