Does the law of total probability apply to hazards? Consider the hazard function for a random variable $T$, conditional on some other random variable $U$:
$$
h(t|U=u)=\lim_{\Delta t\rightarrow0}\frac{P(t<T<t+\Delta t|T>t,U=u)}{\Delta t}
$$
where $U$ has probability density function $f_U(u)$. Suppose, given $h(t|U=u)$, we would like to derive the hazard function irrespective of $U$, what we could call the marginal hazard function:
$$
h(t)=\lim_{\Delta t\rightarrow0}\frac{P(t<T<t+\Delta t|T>t)}{\Delta t},
$$
My question is, can we apply the law of total probability in this setting? In other words, is it the case that $h(t)=E_u[h(t|U=u)]=\int_\mathcal{U}h(t|U=u)f_U(u)du?$
Here is my best attempt to answer the question. I'm not very familiar with measure theory so please forgive me if I get this wrong. Essentially, we are trying to bring a limit outside of a integral, which implies Lebesgue's dominated convergence theorem (DCT) might be useful. DCT states that, for a sequence of functions $f_n(x)$ with $n\in\mathbb{N}$, $\lim_{n\rightarrow\infty}\int f_n(x)dx=\int \lim_{n\rightarrow\infty} f_n(x)dx$, so long as there exists some function $g(x)$ such that $g(x)\geq|f_n(x)|$ for all $x, n$, where $|.|$ indicates absolute value.
To apply DCT, we can re-write the conditional hazard by first defining a sequence of functions (replacing $\Delta t$ with $1/n$):
$$
h_n(t|U=u)=nP(t<T<t+1/n|T>t,U=u),
$$
for $n\in\mathbb{N}$. Then the conditional hazard is defined as the pointwise limit of this sequence:
$$
h(t|U=u)=\lim_{n\rightarrow\infty}h_n(t|U=u)
$$
Then, the DCT indicates that, so long as some function $g(u)$ exists such that $g(u)\geq|h_n(t|U=u)|$ for all $n$ and $u$,
\begin{align}
E_u[h(t|U=u)]
&=\int_\mathcal{U}h(t|U=u)f_U(u)du \\
&=\int_\mathcal{U}\lim_{n\rightarrow\infty}nP(t<T<t+1/n|T>t,U=u)f_U(u)du \\
&=\lim_{n\rightarrow\infty}\int_\mathcal{U}nP(t<T<t+1/n|T>t,U=u)f_U(u)du 
\end{align}
which, by law of total probability, equals
$$
\lim_{n\rightarrow\infty}nP(t<T<t+1/n|T>t)=h(t). 
$$
It seems reasonable to assume that such a $g(u)$ exists - if I understand correctly, this would just require that the conditional hazard is finite.
 A: One case where this implication fails: suppose $U$ is uniformly distributed on the open interval $(0, 1)$, and $T = U$. Let $t = 0$. Then conditional on any particular value $U = u$, the hazard is zero, since $P(T < v) = 0$ for all $v < u$. But the unconditional hazard $h(0) = 1$, since for all $0 \le d \le 1$, $\frac{P(T < d)}{d} = \frac{P(U < d)}{d} = 1$.
Admittedly your dominating function $g$ doesn't exist in this example, since as $u$ goes to zero the conditional hazard goes to infinity near zero. But for $t = 0$ the conditional hazard is never actually infinite.

I think your argument that the implication holds when $g$ exists is promising, with 2 possible gaps.
First, for DCT to apply, $g(u)$ has to dominate not just $h_n(u)$, but also $h_n(u) f_U(u)$. When the pdf of $U$ is unbounded this isn't a trivial difference.
Second, in a formal write-up, it might be worth explaining why the existence of the limit
$$\lim_{n \to \infty} nP(t < T < t + \frac{1}{n} | T > t)$$
implies the existence of the limit $h(t)$. For example, note that the existence of $h(t)$ is not implied by the existence of the limit
$$\lim_{n \to \infty} 2^nP(t < T < t + \frac{1}{2^n} | T > t).$$
