Conditioning on a continuous random variable I have a random variable $N(a)$, which depends on a number $a$, having the property that for all $a \in \mathbb{R}$,
$$P(N(a) \geq 1) = p $$ The example I have in mind is $N(a)$ is $T-a$ where $T$ the time of first arrival in a Poisson process after $a$, which is why there is no dependence of $P(N(a) \geq 1)$ on $a$. However, let us not assume anything like this - $N(a)$ is just a random variable for each $a$. 
Let $Z$ be a continuous random variable independent of all $N(a)$. I would like to assert that 
$$P(N(Z) \geq 1) = p.$$ 
My question is how I might justify this. 
It is natural to try to justify it by writing
$$P(N(Z) \geq 1) = \int_{-\infty}^{+\infty} P(N(a) \geq 1) f_Z(a) ~ da = p$$
but what I don't know is how the first equality can be justified. If the variables were discrete, this would follow by conditioning, but how does one condition on the event $Z=a$ of probability $0$?
 A: If $f_Z(a)>0$, then the conditional density of $N(Z)$ given $Z=a$ is
$$f_{N(Z)|Z=a}(x) = \frac{f_{N(Z),Z}(x,a)}{f_Z(a)},$$
so
$$\mathbb P(N(Z)\geqslant 1|Z=a) = \int_1^\infty\frac{f_{N(Z),Z}(x,a)}{f_Z(a)}\mathsf dx. $$
Note that the denominator is independent of $x$. So multiplying by $f_Z(a)$ and integrating with respect to $a$, we get
$$\int_{-\infty}^\infty \mathbb P(N(Z)\geqslant 1|Z=a)f_Z(a)\mathsf da = \int_{-\infty}^\infty \int_1^\infty f_{N(Z),Z}(x,a)\mathsf dx\mathsf da.  $$
Since $\mathbb P(N(Z)\geqslant 1|Z=a)=\mathbb P(N(a)\geqslant 1)$, the left-hand side is the same as the integral in your post. Using Fubini's theorem to interchange the order of integration (which is justified since the integrand is nonnegative and the integral is finite, being a probability), we have
$$ 
\begin{align*}
\int_{-\infty}^\infty \int_1^\infty f_{N(Z),Z}(x,a)\mathsf dx\mathsf da &= \int_1^\infty \int_{-\infty}^\infty f_{N(Z),Z}(x,a)\mathsf da\mathsf dx\\
&= \int_1^\infty f_{N(Z)}(x)\mathsf dx\\
&= \mathbb P(N(Z)\geqslant 1),
\end{align*}
 $$
the desired result.
A: Note: $Z:\Omega\mapsto \mathbb{R} \implies N_Z \in (N_a)_{a\in \mathbb{R}}$
Therefore, we can think of $P(N_Z\geq 1)$ as another random variable $Y:=g(Z(\omega))$, so $Y$ is a function of $Z$.
Based on your definition of $N_a$, it seems like $Y=p,a.s.$, since:
$$P(Y\neq p)\equiv P(P(N_Z\geq 1)\neq p), \text{ but } \{a\in \mathbb{R}:P(N_a\geq 1)\neq p\}=\emptyset \implies P(Y\neq p)=P(\emptyset)=0$$
Therefore, $P(N_Z\geq 1)=p, a.s.$
$\square$
