Keep in mind that functional derivatives are taken by varying the function, not its argument. So unless the functional itself involves derivatives, and integration by parts is involved in transforming the result (as in the Euler-Lagrange equations, for example) there is no reason for derivatives of the function to appear. Consider an even better case with continuous $K(x,y)$:
$$Z[\phi]=\iint K(x,y)\phi(x)\phi(y)dxdy$$
Then the second variation is $2K(x,y)\delta\phi(x)\delta\phi(y)$, so the second functional derivative comes out as $2K(x,y)$. The example they give simply takes a more singular kernel $K(x,y)=\delta(x-y)$ so the functional reduces to the diagonal integration of the square:
$$Z[\phi]=\iint \phi^2(x)dx.$$
If we stick some sign-changing kernel $k(x)$ into it,
$$Z[\phi]=\iint k(x)\phi^2(x)dx,$$
the second functional derivative will be a signed Borel measure $2k(x)\delta(x-y)$ supported on the diagonal $x=y$. Indeed, we can take
$$Z[\phi]=\iint \phi(x)\phi(y)\,\mu(dx,dy)$$
as a functional for any signed Borel measure $\mu$.
A signed measure $\mu $ (or even a complex valued measure) defined on the Borel $\sigma $-algebra of a locally compact Hausdorff
space $X$ is said to be regular if the total variation measure $|\mu |$ is regular in the sense of Definition 3.14.
For signed measures this is equivalent to both $\mu _+$ and $\mu _-$, from the Hahn decomposition, being regular while, in the
complex case, this is in turn equivalent to both $\Re(\mu )$ and $\Im(\mu )$ being regular signed measures.
It is interesting to observe that, if one chooses the Baire $\sigma $-algebra instead, namely the $\sigma $-algebra generated by the
compact $G_\delta $-sets, then all measures that take finite values on compact sets are automatically regular, provided $X$ is
$\sigma $-compact (Theorem 27 in Royden-Fitzpatrick).
This is a very compelling reason why we should replace the Borel $\sigma $-algebra by its Baire counterpart whenever doing analysis on
locally compact spaces! In my opinion, there are two reasons why we dont't do it, tradition of course being one of them but, most importantly, the reason is the original sin consisting of the following thought process that must have
taken place sometime in history:
If we want to study continuous functions from the point of view of
measure theory, then we'd better introduce a $\sigma $-algebra
relative to which all continuous functions are measurable, meaning
that we want sets of the form $f^{-1}[a,b]$ to be measurable. Since
these sets are closed, why not take the $\sigma $-algebra generated by
all closed sets?
Well, this looks good, except that sets of the form $f^{-1}[a,b]$ are not only closed, but also $G_\delta $, since
$$
f^{-1}[a,b]= \bigcap_{n\in {\mathbb N}}f^{-1}\Big (a-\frac 1n,b+\frac1n\Big ).
$$
So there is no need to take all closed sets, but only some of them!
The question becomes even more subtle
because, when $X$ is compact metrizable, all closed sets are automatically $G_\delta $, so the Borel and Baire $\sigma $-algebras
coincide. However, this is not so for non-metrizable compact spaces (which were probably not in the radar of
the measure theory pioneers anyway).
The price we pay until today is thus to be burdened by requiring regularity of measures in theorems such as
Riesz-Markov-Kakutani and many others, while we could do away with it altogether simply by choosing the right
$\sigma $-algebra!
Best Answer
$\nu$ takes only finite values. So disjointness of $(A_n)$ implies that $\sum \nu(A_n)$ is a convergent series. So is any rearrangement of terms since disjointness still holds. If a series of real numbers converges whenever the terms are permuted then the series is absolutely convergent. Hence $\sum |\nu (A_n)| <\infty$.