When is an integral transform trace class? Given a measure space $(X, \mu)$ and a measurable integral kernel $k : X \times X \rightarrow \mathbb{C}$, the operator 
$$ K f(\xi) =\int_{X} f(x) k(x,\xi) d \mu(x),$$
the operator $K$ is Hilbert Schmidt iff $k \in L^2(X \times X, \mu \otimes\mu)$! 
Q1:The main point of this questions, what are necessary and sufficient conditions for it to be trace class? 
I know various instances, where
$$ \mathrm{tr} K = \int_X k(x,x) d \mu(x).$$
Q2:What are counterexamples, where $x \mapsto k(x,x)$ is integrable, but the operator is not trace class?
Q3:What are counterexamples for a $\sigma$ finite measure space, where $k$ is compactly supported and continuous, but the kernel transformation is not trace class and the above formula fails?
Q4: Is there a good survey/reference for these questions.
 A: A remark on (Q3):
There is this famous example of T.Carleman (1916 Acta Math link) where he constructs a (normal ) operator with a continuous kernel such that it belongs to all Schatten p-classes if and only if $p\geq 2.$ 
More precisely it's possible to construct $k(x)=\sum_n c_ne^{2\pi i n x}$ continuous and periodic with $\sum_n|c_n|^p=\infty$ for $p<2$. Then $Tf=f\ast k$ acting on $L^2(\mathbb T)$ yields the desired result.
Provided some extra regularity on the kernel, the trace formula works fine (there are a lot of results in the literature)
Regarding (Q4) I personally find C. Brislawn's result very interesting but rather difficult to implement in practice.
A: It may be worth noting the phenomena that can appear in Hilbert spaces, where study of the things is more decisive, both positive and negative.
First, I like the "definition" of "trace class" $T:X\rightarrow Y$ with Hilbert spaces $X,Y$ to be that $T$ is a composition of two Hilbert-Schmidt operators (which are defined as being in the HS-norm completion of the algebraic tensor product $X^*\otimes_{\mathrm {alg}}Y$. This gives an intrinsic definition... which, if desired, is provably equivalent to the (ugly) requirement that $\sum |\langle Tx_i,y_i\rangle| <\infty$ for every pair of orthonormal bases.
The reason I recall this cliche is that, in many applications of interest (to me!), natural operators are visibly Hilbert-Schmidt (if compact at all), and the issue becomes to prove trace-class. In practice (for me) it often happens that we know that every one of these integral operators is a finite sum of compositions of two such, proving trace-class. 
Sometimes proof of the latter is highly non-trivial, as in the Cartier/Dixmier-Malliavin proof that test functions on Lie groups are finite linear combinations of convolutions of pairs of such. The totally-disconnected group analogue is trivial.
That summing or integrating down the diagonal fails is easy to illustrate with not-normal operators: the shift operator on one-sided or two-sided $\ell^2$ might seem to have trace absolutely summing to $0$, but it is not trace class at all. Integral analogues of this are clear.
Edit: in response to question about reference, etc.: in Lang's "SL(2,R)" the equivalence of the coordinate-dependent definition of "trace class", and the definition as composition of two Hilbert-Schmidt, are carefully compared. Further, in that same source, various conditions on a kernel assuring that its trace is equal to its integral over the diagonal are carefully treated. (I must say "... in contrast to dangerously glib treatments elsewhere").
Further edit: in response to Yemon Choi's comments: yes, the space of trace-class operators is also the closure of finite-rank operators with respect to the "trace norm"... At the moment, verification of the equivalence seems straightforward.
A: There are many results of the kind you ask about in the book
I. C. Gohberg and M. G. Krein, Introduction to the theory of linear nonselfadjoint operators. Providence, RI: American Mathematical Society, 1969.
It contains both necessary and sufficient conditions, and counter-examples.
