Strange definitions about basic probability - need clarification In the book Condition. The Geometry of Numerical Algorithms - Peter Burgisser, Felipe Cucker, there is an explanation that doesn't make sense to me. It's just a problem about definitions, but I really need a light on this. This section of the book is supposed to be a review of probability, so it's not necessary to know the book to answer my question, just some good knowledge of probability theory. 
Below is the page from the book which I'm having trouble with.

The first problem comes from the integral $\int_M f = 1$, which makes part of the definition of $f$. This integral has to be defined in respect to some measure over $M$ (more precisely, there is a measure $\mu$ over $M$ so that $\int_M f = \int_M f\ d\mu$) . Are they assuming this definition makes sense for any measure over $M$? Or maybe already there is some probability measure over $M$?  
My second problem is: after the definition of density, they say we can use $f$ to define a probability measure $\mu$ on $M$ with another integral, but this another integral doesn't seen to depend on $f$. Where $f$ is hidden in this definition? And again, we are integrating in respect to what measure?
Finally, this is just an observation, but I need some clarification on this too. Usually we work with sample space $\Omega$ instead data space $M$. I'm assuming this is just a mater of conventions, so this is ok. But as far as I know, usually we work with some random variable $X:\Omega\to S$ ($S$ is a Polish space) and the density is a function $f:S\to[0,\infty]$. The way they are defining the density (before mention random variables) makes it be $f:\Omega\to[0,\infty]$. Honestly, I never saw a density defined directly on the sample space. Is this a normal routine?
Thank you and sorry for the long post. 
 A: You have asked three questions. Hopefully I have got them right:


*

*Is there an implicit measure assumed in the definition of a probability density?

*Why does the definition of a probability measure not reference the probability density?

*Is this a definition of a density over a sample space?
I will now try to answer these one after the other.
Implicit Measure
I agree with Andreas Blass: you are right that there must be some implicit measure. Andreas Blass gives the most sensible interpretation in his comment. I'll just expand on the comment here in a bit more detail. Let's give the author the benefit of the doubt and assume that a "data space" is a measurable space equipped with an explicit measure $\mu$ (Wikipedia calls it a "measure space"- see this link). Let's further assume that any operations over this data space, that normally depend on some measure, are implicitly using the measure $\mu$, even if they don't mention it. E.g. if we are integrating over a "data space" $M$ then we are doing so with respect to its implicit measure $\mu$. As Andreas says, this takes care of your first question.
Alternatively, we could spell things out. Since $M$ is a data space, we're assuming it has an implicit measure $\mu$, by the above. Then by the following statement:
$$\int_M f = 1$$
what is meant is
$$\int_M f\,d\mu = 1$$
Note that this works for any measure, not just a probability measure. So let's keep our assumptions as flexible as possible and interpret the author's "data space" as a measure space (measurable space + measure). We don't have to assume the measure is a probability measure. Furthermore, note that our interpretation here assumes that $f$ is defined in terms of some single measure space. That is, the definition of $f$ is not saying that, for all measures $\rho$, $f$ must satisfy:
$$\int_M f\,d\rho = 1$$
It is just saying that given a measure space $M$, with a single implicit measure $\mu$, $f$ is a probability density over this measure space whenever 
$$\int_M f\,d\mu = 1$$
Where is the Density Function?
OK, your second question got me really alarmed with this author. I am going to stick my neck out here and say that this definition is just plain wrong. I may be wrong though, as this is a published book and I am an unpublished nobody. However, I will soon suggest what I think the author meant. First though, I'll use $\rho$ to denote the probability measure we're defining, as we're already using $\mu$ as the implicit measure of $M$. Keeping things explicit then, I believe what was meant was, $f$ defines a probability measure $\rho$ over $M$ such that:
$$\rho(A) = \int_A f \, d\mu$$
That is, the probability measure over some measurable subset $A$, is the integral over that set $A$ of $f$ with respect to the implicit measure $\mu$. Of course, I could be wrong, but this seems like the correct definition, as opposed to the one in the book.
Update: The commenter bartgol has enlightened me about the charecteristic function of $A$ over $M$, denoted $1_A$, which is $1$ for all $x \in A$ and $0$ elsewhere. So I'd say the author meant:
$$\rho(A) = \int_M 1_A f \, d\mu$$
Which is really just a long-winded way of saying what's above.
Density over a Sample Space- Huh?
I think in this case you may have interpreted things wrongly. Though then again, so may I, being a nobody with zero credentials. I believe the space $M$ that we are dealing with in the first place is not the sample space, but indeed the co-domain of the random variable. You say that this is usually a Polish space. Perhaps, but here I think they're just assuming that it's a measure space a.k.a. a "data space". In other words, the random variable will be of type $X : \Omega \rightarrow M$, for some (hitherto unmentioned) sample space $\Omega$. We are then defining our probability density function $f$ of type $M \rightarrow [0, \infty]$. We hope in practice that $f$ indeed captures the "true" probability density of the underlying random variable, as it is distributed over $M$, but this is not necessary for the definition.
Disclaimer
Any or all of the above may be wrong, but this is how I would have interpreted things, having read them. I validate my understanding using mental models in my head of probability distributions over the real numbers and they seem to agree with my interpretation, but maybe I am foolish to do so. Please keep an open mind and if you are an expert reading this, don't judge me too harshly. :)
A: I would say a density $f$ defines a probability measure $P$ via the integral $\displaystyle P(A) = \int_A f$.
Omitting even to mention $f$ in this statement makes no sense to me.
Where the book says $\displaystyle\int_M \mathbf 1_A$ it should have said $\displaystyle \int_M \mathbf 1_A \cdot f$ or $\displaystyle P(A) = \int_A f$.
To avoid confusion resulting from the use of the same symbol for several different things in the question and the answers, I will use the letter $m$ for a measure on $M$.  The integral above is in fact an integral with respect to a measure $m$ on $M$.  So
$$
P(A) = \int_A f\,dm.
$$
One says that $f$ is the density of the probability measure $P$ with respect to $m$.  The measure $m$ need not be a probability measure.  One can also express this by saying $f = \dfrac{dP}{dm}$, and then one says $f$ is the Radon–Nikodym derivative of $P$ with respect to $m$.
