Why does $(2)$ equal $(3)$ with $\nu=\delta_0$ ($\delta_0$ denotes Dirac delta measure at $0$)?

Question

I quote Kuo (2006):

Let $C$ be the Banach space of real-valued continuous functions $\omega$ on $[0,1]$ with $\omega(0)=0$.
A cylindrical subset $A$ of $C$ is a set of the form $$A=\{\omega\in C: (\omega(t_1),\omega(t_2),\ldots,\omega(t_n))\in U\}\tag{1}$$ where $0<t_1<t_2<\ldots<t_n\leq 1$ and $U\in\mathcal{B}(\mathbb{R}^n)$, the Borel $\sigma$-field.
Let $\mathcal{R}$ be the collection of all cylindrical subsets of $C$. Obviously, $\mathcal{R}$ is a field. However, it is not a $\sigma$-field.
Suppose $A\in\mathcal{R}$ is given by $(1)$. Define $\mu(A)$ by $$\mu(A)=\displaystyle{\int_U \prod_{i=1}^n}\bigg(\frac{1}{\sqrt{2\pi(t_i-t_{i-1})}}\exp\bigg[-\frac{(u_i-u_{i-1})^2}{2(t_i-t_{i-1}))}\bigg]\bigg)du_1\ldots du_n\tag{2}$$ where $t_0=u_0=0$
[...] Now, consider the probability measure on $\mathbb{R}^n$ to be defined as if follows: $$\mu_{t_1,t_2,\ldots,t_n}(U)=\displaystyle{\int_{\mathbb{R}}\int_{U}\ \prod_{i=1}^n}\bigg(\frac{1}{\sqrt{2\pi(t_i-t_{i-1})}}\exp\bigg[-\frac{(u_i-u_{i-1})^2}{2(t_i-t_{i-1}))}\bigg]\bigg)du_1\ldots du_n d\nu(u_0)\tag{3}$$ where $U\in\mathcal{B}(\mathbb{R}^n)$, $\nu$ is a probability measure on $\mathbb{R}$ and we use the following convention for the integrand: $$\displaystyle{\frac{1}{\sqrt{2\pi t_i}}}e^-{\displaystyle{\frac{(u_1-u_{0})}{2t_1}}du_1}\bigg\vert_{t_1=0}=d\delta_{u_0}(u_1)\tag{4}$$ where $\delta_{u_0}$ is the Dirac delta measure at $u_0$.

Observe that the integral in the right-hand side of $(3)$ with $\nu=\delta_0$ is exactly the same as the one in the right-hand side of equation $(2)$ for the Wiener measure $\mu$.
[...] Consider now the stochastic process $$Y(t,\omega)=\omega(t),\text{ }\omega\in\mathbb{R}^{[0,\infty)}$$ If we set $n=1$ and $t_1=0$, by $(3)$ and $(4)$, we have that: $$\mathbb{P}\{Y(0)\in U\}=\displaystyle{\int_{\mathbb{R}}\int_{U}\frac{1}{\sqrt{2\pi t_i}}}e^-{\displaystyle{\frac{(u_1-u_{0})}{2t_1}}du_1}\bigg\vert_{t_1=0}d\nu(u_0)\tag{5}$$ $$\begin{split}=\displaystyle{\int_{\mathbb{R}}\bigg(\displaystyle{\int_U}d\delta_{u_0}(u_1)\bigg)d\nu(u_0)}\end{split}$$ $$\begin{split}=\displaystyle{\int_{\mathbb{R}}\delta_{u_0}(U)d\nu(u_0)}\end{split}$$ $$\begin{split}=\nu(U)\text{, }U\in\mathcal{B}(\mathbb{R})\end{split}$$

Some doubts:

1. Does $(4)$ mean that the "quantity" $\displaystyle{\frac{1}{\sqrt{2\pi t_i}}}e^-{\displaystyle{\frac{(u_1-u_{0})}{2t_1}}du_1}\bigg\vert_{t_1=0}$, evaluated at $t_1=0$, equals $d\delta_{u_0}(u_1)$?;
   
2. Does it hold true that $\delta_{u_0}=\delta_0=1$ by definition?
3. Why "the integral in the right-hand side of $(3)$ with $\nu=\delta_0$ is exactly the same as the one in the right-hand side of equation $(2)$ for the Wiener measure $\mu$"?
4. Why, in the last equality of $(5)$, $\int_{\mathbb{R}}\delta_{u_0}(U)d\nu(u_0)=\nu(U)\text{, }U\in\mathcal{B}(\mathbb{R})$ and NOT $\int_{\mathbb{R}}\delta_{u_0}(U)d\nu(u_0)=\delta_{u_0}(U)\cdot\nu(\mathbb{R})$?

Answer 1

I will use the notation $\nu(\mathrm d u)$ instead of the confusing notation $\mathrm d\nu(u)$ as explained in my answer here

1. (4) is more rigorously the limit when $t_1\to 0$ in the sense of measures (or in the sense of distributions).
2. No, $\delta_c$ is the Dirac delta centered at $c$, which is different from the function $1$. One has $\delta_{u_0} = \delta_0$ if and only if $u_0=0$. However, what is true is $∫ \delta_0 = ∫ \delta_{u_0} = 1$.
3. Because by the definition of $\delta_0$ as a distribution, if $\nu = \delta_0$ for any continuous function $\varphi$, $$ ∫_{\mathbb R}\varphi(u_0)\,\nu(\mathrm{d} u_0) = \varphi(0) $$ and now if you take $\varphi(u_0)$ as the integral in $(2)$, since it is written that $t_0=u_0=0$, $(2)$ is nothing but $\varphi(0)$.
4. $u_0$ is the integrated variable in $(5)$, so in any case it cannot be in the final result! First remark that by definition of $\delta_{u_0}$ as a measure, one has $\delta_{u_0}(U) = \mathbf{1}_{U}(u_0)$ (i.e. $\delta_{u_0}(U) = 1$ if $u_0∈ U$, and $0$ if $u_0∉ U$). Therefore $$ \int_{\mathbb R} \delta_{u_0}(U)\,\nu(\mathrm d u_0) = \int_{\mathbb R} \mathbf{1}_{U}(u_0)\,\nu(\mathrm d u_0) = \int_{U} \,\nu(\mathrm d u_0) = \nu(U) $$