Let T be an absolutely continuous random variable and let U be an arbitrary random variable.
Then why does this equality hold: $P(t<T\leq t+s, T\leq U)=\int_t^{t+s} - \frac{\partial }{\partial v}P(T\geq v, U\geq u)|_{v=u}du$
It is not even clear to me why $P(T\geq v, U\geq u)$ would be differentiable almost everywhere with regard to $v$. I am not used to think about absolute continuity. I know this is probably very trivial so if someone would just point me towards the right theorems that would be great.
If you extract the desired term you will get the following result. $$ P(T \geq v, U \geq u) = 1 − P((T \geq v, U \geq u)^C) = 1-((T \geq v)^C \cup (U \geq u)^C) = 1 - P((T < v) \cup (U<u)) = 1 -[P(T < v) + P(U < u) - P(T < v, U < u)] = 1 - F_T(v) + P_U(u)-F_{T,U}(v, u) $$ You stated that T is absolutely continous, and U is arbitrary continous. Hence the result from above should also be continous.