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My background in math (never took a course on measure theory) is weak, but I have been studying it for probability theory here and there as needed. I like to think I understand what a measure, probability measure, and $\sigma$-finite measure is (as a user), but I am reading a paper on Markov chain Monte Carlo, and in its excerpt on convergence theory, it states the following:

Let $X = (X^0, \dots, X^t, \dots)$, $X^t \in E \subseteq \mathbb{R}^n$ be a Markov chain with transition kernel $K:E\times E \rightarrow \mathbb{R}^n$ such that, with respect to a $\sigma$-finite measure $\nu$ on $\mathbb{R}^n$,

$$P(X^t \in A \mid X^{t-1} = x) = \int_A K(x,x') d\nu(x') + r(x) I[x\in A]$$

If we define $K^{(t)}(x,x') = \int K^{(t-1)}(x,y)K(y,x')d\nu(y)+K^{(t-1)}(x,x')r(x') + \{1-r(x)\}^{t-1}K(x,x')$

then $K^{(t)}(x_0, \cdot)$ is the density (with respect to $\nu$) of $X^t$ given $X^0 = x_0$

This really threw me off. I understood measures as functions that map $\sigma$-algebra (sets) to the non-negative real numbers.

My question is, what does "with respect to a $\sigma$-finite measure $\nu$ on $\mathbb{R}^n$" mean? From the definitions I have studied about measures on $\sigma$-algebra, I do not how a density is defined "with respect to" one.

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  • $\begingroup$ The answers below are valid, but just in case it might be helpful to have a simple example to keep in mind: the Gaussian distribution (measure) $\mu$ which assigns a probability weight $\mu(A) = \int_A \frac{e^{-x^2/2}}{\sqrt{2 \pi}} dx$ to a measurable set $A\subset\mathbb{R}$ has a density with respect to the Lebesgue measure. This density is $f(x) = \frac{e^{-x^2/2}}{\sqrt{2\pi}}.$ A density naturally induces a measure. $\endgroup$
    – snar
    Jun 15 '19 at 23:38
  • $\begingroup$ @snar keeping the Gaussian measure in mind when studying these definitions really helped put things in context, thanks! $\endgroup$
    – user228809
    Jun 16 '19 at 16:35
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In general, if $(S, \mathcal{S}, \nu)$ is a measure space, $\mu$ is another measure on $(S, \mathcal{S})$, and $f : S \to [0,\infty]$ is an $\mathcal{S}$-measurable function, we say that $f$ is the density of $\mu$ with respect to $\nu$ if, for every set $A \in \mathcal{S}$, we have $$\mu(A) = \int_A f\,d\nu.$$ (If such $f$ exists then it is unique up to $\nu$-a.e. equality.) We write $f = \frac{d\mu}{d\nu}$. The function $f$ is also called the Radon-Nikodym derivative of $\mu$ with respect to $\nu$.

An important necessary and sufficient condition for the existence of a density is given by the Radon-Nikodym theorem:

Let $\mu, \nu$ be two $\sigma$-finite measures on $(S, \mathcal{S})$. Suppose that for every $A \in \mathcal{S}$ with $\nu(A) = 0$, we have $\mu(A) = 0$. (We say that $\mu$ is absolutely continuous with respect to $\nu$.) Then there exists an $f : S \to [0,\infty]$ which is the density of $\mu$ with respect to $\nu$.

Note that the converse is trivial.


In this context, the measure $\mu$ is taken to be the distribution of $X_t$, and the space $(S,\mathcal{S})$ is $\mathbb{R}^n$ with its Borel $\sigma$-algebra. Then the assertion is that for all Borel sets $A \subset \mathbb{R}^n$, we have $$P(X_t \in A) = \int_A K^t(x_0, x) \nu(dx).$$

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  • $\begingroup$ This explanation coupled with the example provided by @snar to put it in context of probability theory was very helpful. Thank you both. $\endgroup$
    – user228809
    Jun 16 '19 at 16:33
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$\sigma$-finite is a particular kind of measure that is the countable union of measurable sets with finite measure.

The other part of the density is about the Radon-Nikodym Theorem. One definition of a probability density function is as the Radon-Nikodym derivative of the induced measure with respect to a base measure, which is what is talked about in your example.

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  • $\begingroup$ I think the Radon-Nikodym theorem is what I need to understand, this should give me a direction to explore. Thank you for this. $\endgroup$
    – user228809
    Jun 15 '19 at 21:48

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