Let $(A,B,C)$ be a random vector with standard uniform margins and distribution function given by $F_{A,B,C}(a,b,c)=\max(a+b-1,0)\min(0.5,c) + \min(a,b)[c-\min(0.5,c)]$ where $(a,b,c)\in[0,1]^3$. Note that this distribution is not absolutely continuous w.r.t. the Lebesgue measure.
I want to compute the following expectation $E[g(A,B,C)|A=a,B=b]$ where $g$ is a measurable function. Because $F$ does not have a density I can not use $\int g(a,b,c)f_{C|A,B}(c|a,b)dc$.
Moreover, I am also wondering how $F_{C|A,B}$ should be defined in this case?
Update: If we approximate $\min(a,b)$ by a parametric copula $C_{\theta}(a,b)$ that has a density and with $\lim_{\theta\to\infty} C_{\theta}(a,b)= \min(a,b)$ and $\max(a+b-1,0)$ by $D_{\theta}(a,b) = b-C_{\theta}(1-a,b)$, so that $\lim_{\theta\to\infty} D_{\theta}(a,b)= \max(a+b-1,0)$, then the resulting conditional density $f_{C|A,B;\theta}$ is well defined and
$\lim_{\theta\to\infty} f_{C|A,B;\theta}(c|a,b)= 2 1\!\!1\{(a,b)\in Q_1\cup Q_3\} +
2 1\!\!1\{(a,b)\in Q_2\cup Q_4\}$, where $Q_i$ denotes the $i$-th quadrant of the unit cube, e.g., $Q_3 = (0,0.5)\times (0,0.5)$.
That is, if we approximate the singular distribution by an absolutely continuous distribution we could conclude (?) that we can compute the expectation using
$\int g(a,b,c)f_{C|A,B}(c|a,b)dc$ with $
f_{C|A,B}(c|a,b)= 2 1\!\!1\{(a,b)\in Q_1\cup Q_3\} +
2 1\!\!1\{(a,b)\in Q_2\cup Q_4\}$?
Is this a valid/common definition of the conditional density/expectation in such a case?
The resulting conditional cdf is given by $F_{C|A,B}(c|a,b) = \frac{1}{\gamma}\min(c,\gamma) 1\!\!1\{(a,b)\in Q_1 \cup Q_3\} + \frac{1}{1-\gamma} (z-\min(\gamma,c))1\!\!1\{(a,b)\in Q_2 \cup Q_4\}$. By definition, we have that $F_{C|A,B}$ is the conditional cdf of $C$ given $A,B$ if for all $(a,b,z)\in[0,1]^3$ it holds that \begin{align*} F_{A,C,B}(a,c,b) & = \int_0^b\int_0^a F_{C|A,B}(c|t,s)dC_{A,B}(t,s). \end{align*} This can readily be verified.