A narrow probability curve defines its space-time measure

Let $G\subseteq {\mathbb{R}}^d$ be open and let $t_n\to t$. Narrow continuity gives ${\mu }_{t_n}\Rightarrow {\mu }_t$, so [F2] gives ${\mu }_t(G)\le {\mathrm{lim\hspace{0.17em}inf}}_n{\mu }_{t_n}(G)$. Thus $t\mapsto {\mu }_t(G)$ is lower semicontinuous and hence Borel measurable.

Let $\mathcal{D}$ be the Borel sets $A\subseteq {\mathbb{R}}^d$ for which $t\mapsto {\mu }_t(A)$ is measurable. It contains the whole space because each section is a probability. If $A\in \mathcal{D}$, then ${\mu }_t(A^c)=1-{\mu }_t(A)$, so $A^c\in \mathcal{D}$. For pairwise disjoint $A_j\in \mathcal{D}$, countable additivity gives ${\mu }_t({\bigcup }_j{A}_j)={\lim }_{N\to \mathrm{\infty }}{\sum }_{j=0}^N{\mu }_t(A_j)$, a measurable increasing limit. Hence $\mathcal{D}$ is a lambda-system. The open sets form a pi-system generating the Borel sigma-algebra and belong to $\mathcal{D}$ by step 1.1, so [F3] and [F9] give $\mathcal{D}=\mathcal{B}({\mathbb{R}}^d)$. Together with [F4], this proves that $t\mapsto {\mu }_t$ is a probability kernel.

By [F5], a product-Borel set $E$ is measurable for $\mathcal{B}([0,T])\otimes \mathcal{B}({\mathbb{R}}^d)$. Its sections $E_t$ are Borel by [F6]. Since $t\mapsto {\mu }_t$ is a probability kernel by step 2.1, [F7] applied to ${\mathbf{1}}_E$ shows that $t\mapsto {\mu }_t(E_t)$ is measurable. Thus $M(E)={\int }_0^T{\mu }_t(E_t)dt$ is defined; $M(\mathrm{\varnothing })=0$ and $M([0,T]\times {\mathbb{R}}^d)={\int }_0^{T}1dt=T$.

If $E_j$ are disjoint product-Borel sets, their sections are disjoint and ${\mu }_t(({\bigcup }_j{E}_j{)}_t)={\sum }_j{\mu }_t((E_j{)}_t)$. Integrating the increasing partial sums and using [F8] gives $M({\bigcup }_j{E}_j)={\sum }_{j}M(E_j)$; with step 2.1 this makes $M$ a finite Borel measure. For an indicator, the iterated-integral formula is the definition of $M$; finite nonnegative simple combinations give it for simple functions. For arbitrary nonnegative product-Borel $f$, use [F9] to choose $s_n\uparrow f$ and apply [F8] to both $M$ and the time-section integrals. This proves $\int fdM={\int }_0^T\int f(t,x)d{\mu }_t(x)dt$.