 Solution

Assume $h(x)>0$, so the conditioning event has positive probability. For a continuous-time Markov chain, write ${\tau }_y={\int }_0^{T_A}{1}_{\{X_s=y\}}ds$. At each time $s<T_A$, the Markov property gives $\mathbb{P}(T_A<\infty \mid {\mathcal{F}}_s)=h(X_s)$. Using Tonelli theorem for the nonnegative occupation integral,Division by $h(x)$ yields $\boxed{{\mathbb{E}}_x[{\tau }_y\mid T_A<\infty ]=[h(y)/h(x)]{\mathbb{E}}_x{\tau }_y}$. The discrete-time proof replaces the integral by a sum. Infinite expectations are allowed in the nonnegative sense; if $h(y)=0$, the numerator is zero directly, avoiding an ambiguous product of zero and infinity.