Would you accept a trace-integral as a double sum? The distribution character of a representation $\pi$ (of a, let's say semisimple, $p$-adic group $G$, although I think that most of what I say here is also true for real Lie groups) is given by $$ \operatorname{tr} \pi(f) = \operatorname{tr} \int_G \pi(g)f(g)\mathrm dg. $$ One would very much like to say, as in the case of $G$ compact, that $$ \operatorname{tr} \pi(f) = \int_G (\operatorname{tr} \pi(g))f(g)\mathrm dg, $$ but, in almost all interesting cases. Although this is not possible for $\pi$ infinite-dimensional unitary, it is a very deep result of Harish-Chandra that there is a locally integrable function $\Theta_\pi$ with various nice properties (for example, it is locally constant on the regular semisimple locus) so that $$ \operatorname{tr} \pi(f) = \int_G \Theta_\pi(g)f(g)\mathrm dg. $$ The computation of this function $\Theta_\pi$ is, due to its indirect definition, quite difficult. My own research is largely concerned with the case where $\pi$ is an irreducibly compactly induced representation $\pi = \operatorname{Ind}_K^G \rho$, in which case more work of Harish-Chandra shows that $$ \Theta_\pi(\gamma) = \int_G \int_K \dot\theta_\rho(g k\gamma k^{-1}g^{-1})\mathrm dk\,\mathrm dg. $$ (The inner integral over a compact open subgroup $K$ also comes, if one digs into the details of the proof, from the failure of a double integral to be ‘switchable’.)
I believe that the (non-compact case of) the trace formula can also be viewed as a lengthy and difficult matter of dealing with the failure of Fubini, but I couldn't talk in any intelligible fashion about this (and could well be wrong). Perhaps, if I'm not completely off, someone else could discuss it.