Baaquie, in "Quantum Finance", states that the Dirac Delta function is unmeasurable, since it "has support on a set that has zero measure"
What is a "support"? What kind of mathematical object is it (e.g. function, set, relation, point...)? Why does having support on a set with zero measure makes it Lebesgue-unmeasurable?
I think that part of your confusion comes from the fact that the action of the $\delta$-distribution (functional) on a test function is written by the use of the integral sign. In fact, you should think of this integral $$\int\limits_{-\infty}^{+\infty}{dx f(x)\delta (x-a)}$$ just as a symbol (a notation). It is not a real integral and the integrants are not real functions ($\delta(x-a)$). To avoid confusion, it should be written with the usual notation for duality product: $\langle \delta(x-a),f\rangle=\langle \delta_a,f\rangle =f(a)$.
Many distributions can be indeed represented by an integral and a Lebesgue measurable function: If $\phi$ is a distribution and $g\in L_{loc}^1(\mathbb R)$, for which $\langle \phi, f\rangle=\int\limits_{-\infty}^{+\infty}{g(x)f(x)dx}$, for all test functions $f$ then the distribution $\phi$ is identified with the locally integrable function $g$ (in fact every $g\in L_{loc}^1(\mathbb R)$ defines a distribution by the above integral) and often it is used the same letter for the distribution and the function, i.e often it will be written like $\langle \phi, f\rangle=\int\limits_{-\infty}^{+\infty}{\phi(x)f(x)dx}$. Such distributions, which can be represented by the above integral are called regular. It is well known that the $\delta$ distribution is not regular, i.e it can not be expressed in terms of integral with the use of some locally integrable function $g$. But still many authors prefer to use the integral representation like for the regular distributions in order to keep an unified notation and way of exposition. Therefore, the integral $\int\limits_{-\infty}^{+\infty}{dx f(x)\delta (x-a)}$ is just a symbol, and is understood as if there was really a locally integrable function $\delta(x-a)$ for which $\langle \delta(x-a),f\rangle= \int\limits_{-\infty}^{+\infty}{dx f(x)\delta (x-a)}=f(a)$ (where we again use the same letter for the distribution and the "$L_{loc}^1$-function").
A good reference, that I recommend you to check is "Green functions and boundary value problems" (3-rd edition) by Ivar Stakgold and Michael Holst.
Support for a normal function is the region where the function take non-zero values (or the closure of that set to be precise).
Now for the dirac $\delta$ it's a generalized function or distribution. They are a special kind of mapping from infinitely derivable functions with compact support (test function) to real numbers. One can in a simplified way say that $\int \delta\varphi\,dx$ is defined for each test function $\varphi$. This means that a distribution is not defined in terms of the pointwise values it takes.
Now again for the support of a distribution $f$. That's a set such that every test function that is zero there we will have $\int f\varphi\, dx=0$.
But the talk about lebesgue measurable is not quite correct the way you interpret it. They're talking about it not being a ordinary lebesgue measurable function which is true since it's not a function to start with. It's meaningful to talk about the integral of a distribution under certain circumstanses. The point of it being no ordinary function is the point because $\int \delta dx=1$, but $\int_{\operatorname{supp}\delta}dx=0$ which would be a contradiction for a function.
