What does $\ll$ mean in the context of measure theory? I don’t know much about measure theory, but I have seen the notation $P\ll Q$, where $P$ and $Q$ are measures, appear a few times. Unfortunately, I haven’t seen a formal definition (it’s also something that is hard to plug into Google!). Can anyone point me in the right direction?
An interesting follow-up that has emerged in the comments is why this notation?
 A: $P\ll Q$ means $P$ does not assign positive measure to any set to which $Q$ does not assign positive measure.
An obvious case in which this holds is when $P$ is defined as follows:
$$
P(A) = \int_A f\,dQ.
$$
In that case the function $f$ is called the density of $P$ with respect to $Q$ or the Radon–Nikodym derivative of $P$ with respect to $Q.$ Sometimes one writes
$$
f= \frac{dP}{dQ}.
$$
What textbooks report to be "probability density functions" are Radon–Nikodym derivatives of probability measures on $\mathbb R^n$ with respect to Lebesgue measure on $\mathbb R^n.$
An obvious case of lack of absolute continuity is when $P$ assigns positive measure to a set containing just one point in $\mathbb R.$ Thus if you throw a die $n$ times, with probability $1/6$ of getting a $1$ each time, the probability that the number of $1\text{s}$ is $k$ is $\binom n k \left( \frac 1 6\right)^k\left(1  - \frac 1 6\right)^{n-k},$ and there is no function $f$ for which, for every Lebesgue-measurable $A\subseteq\mathbb R,$
$$
\int_A f\,dm = \sum_{k\,:\,k\,\in\,A} \binom n k \left( \frac 1 6\right)^k\left(1  - \frac 1 6\right)^{n-k}
$$
(where $m$ is Lebesgue measure on the line).
A less obvious case is this: Let the $n$th base-three digit of a number $X$ be $0$ or $2$ according as one gets tails or heads on the $n$th independent toss of a "fair" coin.
Then we have $\Pr(X\in[0,1/3]\cup[2/3,1]) = 1$ and for $n=1,2,3,\ldots,$
then constraint that the $(n+1)$th digit is $0$ or $2$ requires $X$ to lie in a set whose Lebesgue measure is $2/3$ that of the set to which $X$ is constrained to lie by the allowed values of the first $n$ digits. Let
$$
C= \{x\in\mathbb R: \forall\varepsilon>0\,\, \Pr(X\in(x\pm\varepsilon)>0\}.
$$
Then we have shown that the Lebesgue measure of $C$ is no more than $(2/3)^n$ for $n=1,2,3,\ldots,$ and so we conclude that $m(C)=0.$
So

*

*$\Pr(X\in C) = 1$ and $m(C)=0$ and

*for every $x\in\mathbb R,$ $\Pr(X=x)=0.$
The condition after the last bullet point means that the probability distribution of $X$ has no discrete part; hence the cumulative probability distribution function $x\mapsto\Pr(X\le x)$ is continuous.
So here we have a probability measure on Borel subsets of $\mathbb R$ that assigns all of the probability to a set whose Lebesgue measure is $0$ despite assigning probability $0$ to every point.
This is not absolutely continuous with respect to Lebesgue measure.
