Understanding operator valued distributions Suppose $a(x)$ is an operator valued distribution (i.e. a linear map that associates an operator to each test function). If $f$ is a test function in a Schwartz space, the annihilation operator $A$ is defined as
$$A(f) = \int f^*(x)a(x) dx$$
for some operator-valued distribution $a$.
How do we know that all operator valued distributions can be expressed as an integral, is this just an abuse of notation?
If this is really true integration being carried out, I am having some trouble interpreting the above integral. What exactly does the above integral mean? How does it make sense to integrate a test function against an operator, which I assume is with respect to the Lebesgue measure?
 A: This is a common abuse of notation when using distributions in Physics in general, so not something specific to operator-valued ones. Given a test function $f(x)$ you can always define a distribution which acts on other test functions by integrating them against $f(x)$. These distributions, which are in fact just test functions, are sometimes called regular.
The fact is that not all distributions are regular. The Dirac delta distribution centered at $a\in \mathbb{R}^d$ is defined to be the evaluation functional: $$\delta_a[f]=f(a)\tag{1}$$
but as a notation everyone writes it as $\delta^{(d)}(x-a)$ and its action is represeted by an integral $$\int d^dx \delta^{(d)}(x-a)f(x)=f(a)\tag{2}.$$
It is a fact that there is no true function which realizes (2). For this reason the Dirac delta is not a regular distribution.
Nevertheless, since regular distributions are integrals against functions, the usual practice in Physics is to denote a distribution as if it were a function, say $\xi(x)$, such that its action on test functions is given by an integral. This is just notation: distributions which are not regular only make sense under the integral sign, meaning, when they act on test functions.
As for the other question: what means to integrate a function against an operator. This is a separate question altogether, because now you are talking about true integration and not just a notation for the application of a linear functional. But the basic point is that if you have one operator ${\cal O}$ in a Hilbert space ${\cal H}$, such an operator is determined by its matrix elements $\langle \psi|{\cal O}|\phi\rangle$, since these will determine, for example, the operator action in some orthonormal basis.
In that case you may define the integral of ${\cal O}(x)$ against some function $f(x)$ by $$\bigg\langle\psi\bigg|\int d^dx f(x){\cal O}(x)\bigg|\phi\bigg\rangle=\int d^dx f(x)\langle \psi|{\cal O}(x)|\phi\rangle\tag{3}$$
