Below is from Liptser, Shiryaev "Theory of martingales", page 200:
I have a question: How from eq. 3.5 and 3.8 they got the eq. 3.9?
Since $G$ is of finite variation $\mathcal{E}^{-1}(G)$ is of finite variation too. So I applied integration by parts: $$Z_t=\int^t_0\mathcal{E}^{-1}_{s-}(G)dL_s+\int^t_0 [L_{s-}+\Delta L] d\mathcal{E}^{-1}_s(G)$$ $$=\int^t_0Z_{s-}dG_s+\int^t_0Z_{s-}dX^c_s+\int^t_0\int_{R_0}Z_{s-}(e^{I\lambda x}-1)d(\mu-\nu)_s$$ $$-\int^t_0Z_{s-}\frac{dG_s}{1+\Delta G_s}-\int^t_0Z_{s-}\frac{\Delta G_s dG_s}{1+\Delta G_s}- \int^t_0\int_{R_0}Z_{s-}(e^{I\lambda x}-1)(\mu-\nu)({s},dx)\frac{dG_s}{1+\Delta G_s}$$ $$=\int^t_0Z_{s-}dX^c_s+\int^t_0\int_{R_0}Z_{s-}(e^{I\lambda x}-1)d(\mu-\nu)_s$$ $$-\int^t_0\int_{R_0}Z_{s-}(e^{I\lambda x}-1)(\mu-\nu)({s},dx)\frac{dG_s}{1+\Delta G_s}$$