Why do we write $P\ll Q$ and why is $P$ called absolutely continuous w.r.t. $Q$?

Question

Let $P$ and $Q$ be two measures on a measurable space $(\Omega,\Sigma)$. My book, Stochastic Finance by Föllmer and Schied, says that if $$\forall A\in\Sigma: Q(A)=0\Rightarrow P(A)=0$$ then $P$ is called absolutely continuous w.r.t. $Q$ and we write $P \ll Q$. Why? Obviously continuity already has a clear meaning in mathematics and $x \ll y$ is sometimes informally used to say that $y$ is much larger than $x$. I assume that there are good reasons for this, can someone please elaborate?

Answer 1

The concept was first used by the German Mathematician E.Harnack in $1884$ and it was named by the Italian Mathematician G.Vitali in $1905$. It is not explicitly stated as to why Vitali chose this particular name, although there are some inferences we can make from the properties of absolutely continuous measures.

There are many properties of absolute continuity that are analogous to those of continuity of functions. It's important to remember that because we are talking about measures, the notion of continuity will inherently be different to that of an arbitrary function. Fundamentally, as @LeeMosher points out in the comments, this is just a choice of notation. However, there is some motivation that might help it to make a bit more sense, which I will present below.

Motivation I

As mentioned in my comments, the most fundamental property comes from the Radon Nikodym Theorem. This states that for sigma finite measures $\mu$ and $\nu$:

$$ \nu \ll \mu \iff \frac{d \nu}{d \mu} \space \space \text{exists} $$

Where $\frac{d \nu}{d \mu}$ referse to the Radon Nikodym derivative between the two measures.

Note that this is similar to the principle that applies to arbitrary functions. We know that if a function $f$ is differentiable, then it must be continuous. And this is a similar statement to that, however in this case it is an "if and only if" claim.

And so the fact that we have differentiability between the measures, means that we can say that there is continuity between the measures as an analogue to this.

Motivation II

More generally than I stated in my original comment, the following relation also holds:

The conditions $$ \nu \ll \mu \iff \forall \varepsilon > 0 \space \exists \delta > 0 \space ( \mu(A) \le \delta \;\Rightarrow\; \nu(A) \le \varepsilon ) $$ if it holds that $\mu$ is $\sigma$-finite and $\nu$ is finite.

The proof of this result comes from “An Introduction to Integration and Measure Theory” by Ole Nielsen (Proposition 15.5).

Motivation III

As mentioned in @EricWofsey’s answer here there is a relationship between the absolute continuity of functions and absolute continuity between measures. I hesitate to mention it here as this relationship only exists when specifically considering the Lebesgue measure and does not generalise. I have linked the answer for those interested, however, I don’t believe it answers the question of why the name was chosen since it only really applies in this special case. There is no evidence to suggest that this was the motivation for the name, although it is an interesting observation in spite of this.

Answer 2

There is also a more algebraic perspective, that at least contextualizes the notation it seems to me. Although admittedly there is no new content in what I write below.

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Namely, that given a $\sigma$-algebra of subsets, the collection of its $\sigma$-ideals (an ideal in the sense of algebra (with countable operations allowed) if addition is symmetric difference $A\triangle B = (A\cup B)\setminus (A\cap B)=(A\setminus B)\uplus (B\setminus A)$ and multiplication is intersection) are partially ordered via inclusion. Any measure $\mu$ on a $\sigma$-algebra $\Sigma$ defines a certain $\sigma$-ideal, namely the collection $\mathcal{N}(\mu)$ of $\mu$-negligible subsets. Then

$$\mu\gg \nu \iff \mathcal{N}(\mu)\leq \mathcal{N}(\nu)\,\, (\text{i.e. } \mathcal{N}(\mu)\hookrightarrow \mathcal{N}(\nu) \,).$$

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Continuing along these lines, one can also get at absolute continuity as a sort of continuity. For simplicity let's consider only probability measures. Then $d_\mu(A,B)=\mu(A\triangle B)$ defines a pseudometric on $\Sigma$. Note that the sets that the pseudometric $d_\mu$ does not distinguish are precisely the sets that differ by a $\mu$-negligible set, and the quotient $\sigma$-algebra $\Sigma/\mathcal{N}(\mu)$ becomes a complete metric space with the induced metric $(A\triangle \mathcal{N}(\mu), B\triangle\mathcal{N}(\mu))\mapsto d_\mu(A,B)$ (see the Fremlin reference I've given in my answer at Equivalent definition of weakly mixing for details).

Let us consider an anonymous probability measure $\nu:\Sigma\to[0,1]$. Note that $\nu\ll \mu$ iff $\nu$ factors through $\Sigma/\mathcal{N}(\mu)$. The $\epsilon-\delta$ characterization of absolute continuity mentioned by FD_bfa then says that $\nu\ll \mu$ iff $\nu$ is (uniformly) continuous w/r/t the topology induced by the pseudometric $d_\mu$.

(Alternatively one can compare the topologies on $\Sigma$ induced by $d_\mu$ and $d_\nu$.)