# What does it mean for a transition kernel to be a density with respect to a sigma-finite measure?

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.

• 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.
– snar
Jun 15 '19 at 23:38
• @snar keeping the Gaussian measure in mind when studying these definitions really helped put things in context, thanks! Jun 16 '19 at 16:35

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 $$P(X_t \in A) = \int_A K^t(x_0, x) \nu(dx).$$

• This explanation coupled with the example provided by @snar to put it in context of probability theory was very helpful. Thank you both. Jun 16 '19 at 16:33

$$\sigma$$-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.

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