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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?

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  • $\begingroup$ Absolute continuity between a measure and the Lebesgue measure is equivalent to the following statement: for every $\epsilon > 0$ there exists $\delta > 0$ such that $\mu(A)< \epsilon$ for all Borel sets $A$ such that $Leb(A)<\delta$ $\endgroup$
    – FD_bfa
    Commented Jan 21, 2023 at 15:56
  • $\begingroup$ Also note that absolute continuity for two arbitrary $\sigma$-finite measures is guaranteed if and only if the Randon Nikodym derivative exists between the two measures (Radon Nikodym Theorem) $\endgroup$
    – FD_bfa
    Commented Jan 21, 2023 at 15:59
  • $\begingroup$ @FD_bfa Given that we write $P \ll Q$ I would have expected that e.g. $$\forall A\in\Sigma:P(A)\leq Q(A)$$but as far as I understand your first comment there is no guarantee for this, right? $\endgroup$
    – Filippo
    Commented Jan 21, 2023 at 16:03
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    $\begingroup$ See math.stackexchange.com/a/1695281/86856 for the origin of the name. $\endgroup$ Commented Jan 21, 2023 at 16:33
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    $\begingroup$ As $P\ll P$, this relation is not so much like “much less than” $\endgroup$ Commented Jan 22, 2023 at 21:15

2 Answers 2

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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.

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    $\begingroup$ Maybe a bit of clarification for the OP: A function $f : [a,b]\to \mathbb{R}$ is absolutely continuous if and only if there is an integrable function $g$ such that $f(x) = \int_a^x g(y) dy$, and it turns out that $g = \frac{df}{dx}$ almost everywhere. By the Radon Nikodym theorem, a measure $\mu$ is absolutely continuous with respect to $\nu$ if and only if there exists an integrable function $g$ such that $\mu(A) = \int_A g d\nu$, and in this case we call $g$ the Radon-Nikodym derivative and use the notation $g=\frac{d\mu}{d\nu}$. A priori there’s no differentiability, it’s just an analogy. $\endgroup$ Commented Jan 21, 2023 at 16:31
  • $\begingroup$ (+1) because the answer is useful. I have not accepted it, because I feel like there is an important point missing: As Eric Wofsey explains in his answer, there are relevant cases where $P\ll Q$ is equivalent to a certain function being absolutely continuous. I think that his answer can be generalized by considering Stieltjes integrals. It seems very plausible to me that this influenced the choice of Vitali and in any case it is a good justification. $\endgroup$
    – Filippo
    Commented Jan 22, 2023 at 10:52
  • $\begingroup$ Unfortunately, I’m not sure we can generalise the answer in this way. A necessary and sufficient condition for a finite measure $\mu$ on $\mathbb{R}$ to be absolutely continuous with respect to the Lebesgue measure is that the function $\mu((-\infty, x])$ should be an absolutely continuous function (as described in Eric Wofsey’s answer). However, on the measurable space of the $\mathbb{R}^n$ and the associated Borel sigma algebra, I believe this only holds for the Lebesgue measure (and some trivial measures) @Filippo $\endgroup$
    – FD_bfa
    Commented Jan 22, 2023 at 12:02
  • $\begingroup$ (continued) The reason why I excluded Eric Wofsey's point from my answer was because it really only describes a relationship that exists when consider the Lebesgue Measure, and it doesn't generalise very well. Perhaps this is what inspired Vitali, but I'm not sure it's the most useful way to think about it because it doesn't really give you any properties that apply more generally @Filippo $\endgroup$
    – FD_bfa
    Commented Jan 22, 2023 at 12:09
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    $\begingroup$ Is the bibliographical information from mathshistory.st-andrews.ac.uk/Miller/mathword/a? If so, it might be good to add references. $\endgroup$
    – Alp Uzman
    Commented Jan 23, 2023 at 18:32
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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.


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

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


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 $d_\nu$.)

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  • $\begingroup$ Thank you for your answer (+1). Unfortunately I don't know how $\uplus$ is defined, could you please elaborate? $\endgroup$
    – Filippo
    Commented Jan 23, 2023 at 20:28
  • $\begingroup$ @Filippo It means the union of two sets that also happen to be disjoint. $\endgroup$
    – Alp Uzman
    Commented Jan 23, 2023 at 20:30
  • $\begingroup$ Thank you. Another notation is new to me: $\mathcal{N}(\mu)\hookrightarrow \mathcal{N}(\nu)$. Is this equivalent to $\mathcal{N}(\mu)\subseteq \mathcal{N}(\nu)$? $\endgroup$
    – Filippo
    Commented Jan 23, 2023 at 20:35
  • $\begingroup$ Oh and thank your for editing the title. $\endgroup$
    – Filippo
    Commented Jan 23, 2023 at 20:41
  • $\begingroup$ @Filippo Yes, and no problem. $\endgroup$
    – Alp Uzman
    Commented Jan 23, 2023 at 21:06

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