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.