To be more clear about why I can't make sense of your example, surely your measure function
`\mu(x) is the largest integer less than or equal to x'
doesn't obey the countable additivity requirement \mu\left(\cup_{i\in\Omega}A_i\right)=\sum_{i\in \Omega}\mu(A_i)\;\;\forall A_i\cap A_j=\emptyset...
Let's forget intervals for now, that was just to make the two distinct with notation; I understand that it is is for measurable sets allowing the discussion of your example or the Dirichlet function or what have you: this surely is a result of being able to consider it as a sum over intervals in...
Hi all,
Sorry if this is in the wrong place. I'm trying to understand probability theory a bit more rigorously and so am coming up against things like lebesgue integration and measure theory etc and have a couple of points I haven't quite got my head around.
So starting from the basics...
Hi all,
Apologies if this is stupid question, but I have the following situation. Given two measures u(x) and v(x) if u(x) is absolutely continuous to v(x) ( u<<v) I have a result such that
\int_A f(x)dv(x) always takes the value \int_B g(x)du(x)
But strictly \int_B g(x)du(x)\neq\int_A...