Let $H$ be the Heaviside step function $H:t\in \mathbb{R}\mapsto \mathbb{R}\ni H(t)$.
The measure $$ \nu(dt,dx) = \delta_0(dx) I_{]-\infty,0]}(dt) + \delta_1(dx)I_{[0,+\infty[}(dt), $$ where $I$ is the characteristic function and $\delta$ is the Dirac generalized function, is supported on the graph of $H$ and is called an occupation measure for $H$. The notation $\nu(dt,dx)$ stands for the measure $d\nu(t,x)$ so that $\delta_0(dx)$ is in fact the measure (and not the generalized function) $\delta_0(x)dx$. Then, for two Borel sets $A,B$, $$ \nu(A\times B) = \int_{A\times B} d\nu= \int_{A\times B} \delta_0(x)I_{]-\infty,0]}(t)\;dtdx $$
Now, consider the function $$ f(t) = t, \qquad t\in [0,1]. $$
- Can we write an occupation measure for $f$ as a product between a measure depending on $t$ only and a measure depending on $x$ only ?
From the general viewpoint
Consider the measure $\nu$ supported on a trajectory : $[0,T] \ni t\mapsto x(t) \in X\subset \mathbb{R}$.
Given tow Borel sets $A\subset [0,T],B\subset X$, we can measure the time spent by the trajectory in $B$ during a duration $A$ by computing $$ \nu(A\times B) = \int_A I_{B}(x(t))\; dt \qquad (\dagger). $$
Applying the disintegration theorem to $\nu$ between the sets $[0,T]\times X$ and $T$ (wiki source) , we have $$ \nu(A\times B) = \int_A \nu(B\mid t)\, \nu(\pi^{-1}(dt)) $$ where
- $\pi$ is the canonical projection on $[0,T]$,
- $\nu(B\mid t)$ is the condition expectation : for each $t$, this is a measure on the fiber $\pi^{-1}(\{t\})$, so that $\nu(B\mid t) = \nu(([0,T]\times B)\cap \pi^{-1}(\{t\})$,
- $\nu(\pi^{-1}(dt))$ should be equal to $dt$ since $\pi^{-1}(dt)$ is an infinitesimal area w.r.t to $t$ and we measure the time with the Lebesgue measure?
The theorem says that such decomposition is unique and that $\nu(.\mid t)$ is a probability measure on $X$. Regarding $(\dagger)$, $\nu(B\mid t)=I_{B}(x(t))$ and $I_{.}(x(t))$ is a probability measure on $X$.
Then to answer the first question, we write $$ \nu(A\times B) = \int_A I_B(t) \; dt $$ and we would like to find $d\nu(t,x)$ such that $$ \int_A I_B(t) \; dt = \int_{A\times B}d\nu(t,x). $$ We have $$ \int_A I_B(t) \; dt = \int_A \delta_t(B)\, dt = \int_A \left(\int_B d\delta_t(x)\right) dt $$ so $$ d\nu(t,x) = d\delta_t(x)\, dt. $$