Q: How does one construct a non continuous representation $\rho:SL_2(\mathbf{R})\rightarrow G$ for some connected (finite dimensional) Lie group $G$?

Example can be found, for instance, in Boris Weisfeiler's paper "Abstract homomorphisms of big subgroups of algebraic groups", pages 149150, see http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.ndml/1175197662 His example of a discontinuous representation $\rho$ of $SO(n, {\mathbb R})$ to a semidirect product $H$ of $SO(n, {\mathbb R})$ with the abelian group ${\mathbb R}^N$ (the Lie algebra of $SO(n)$), works for $SL(2, {\mathbb R})$ as well. Actually, Weisfeiler's example is even more dramatic: The image of the compact group $SO(n)$ under $\rho$ is dense in the noncompact Lie group $H$. Weisfeiler's paper also lists many positive results on rigidity of abstract homomorphisms of Lie groups. 


There are natural function spaces on Lie groups that are nevertheless not continuous (and, thus, are not representations in any usual, useful sense). For example, already on $G=\mathbb R$, the Frechet space $V$ of all continuous functions, and/or the Frechet space of bounded continuous functions, with the translation action of $G$, are not repn spaces, in the sense that $G\times V\rightarrow V$ is not continuous. The reason is the existence of notuniformlycontinuous continuous functions. For example, $f(x)=\sin(x^2)$. 


A partial answer: No measurable constructions are possible. Any measurable group homomorphism between locally compact groups is automatically continuous, in fact $C^\infty$ for Lie groups. You can have a look at the answers in an old question of mine: Are measurable automorphism of a locally compact group topological automorphisms? If I would like to find something non continuous, I personally would start with finding some non measurable automorphism of the circle first. 

