Let $Z(t)$ and $E_N^0(f(Z(0)))$ be given as in the snippet below. I would like to know how follows this: $$E[\lambda(0)\ | \ Z(0)]=E[\lambda(0)]$$ from the a.s. uniqueness of the conditional expectation and from this: $$E_N^0[f(Z(0))] = E[f(Z(0))]$$
for all non-negative measurable functions $f:E\to \mathbb{R}$
For random variables $X:\Omega\rightarrow\mathbb{R}$ and $Y:\Omega\rightarrow\mathbb{R}$, we can form a new random variable $E[X|Y]:\Omega \rightarrow\mathbb{R}$, called the conditional expectation of $X$ given $Y$ (also called the conditional expectation of $X$ given the sigma algebra $\sigma(Y)$). For details, you can google "conditional expectation given a sigma algebra."
The conditional expectation $E[X|Y]$ is always a deterministic function of $Y$, so that $$ E[X|Y] = g(Y)$$ for some deterministic and measurable function $g:\mathbb{R}\rightarrow\mathbb{R}$. Since all random variables $X, Y$ are really functions of the outcome $\omega \in \Omega$, we can emphasize such dependency by writing $X(\omega)$ and $Y(\omega)$ and $$ E[X|Y](\omega) = g(Y(\omega))$$
It turns out that there can be infinitely many versions of the conditional expectation $E[X|Y]$. So we can have three different versions $g_1(Y)$, $g_2(Y)$, $g_3(Y)$, all valid versions of $E[X|Y]$. So then what do we mean when we write $E[X|Y]$? Which version do we use: $g_1(Y)$, or $g_2(Y)$, or $g_3(Y)$, or some other version? It turns out that it does not matter since it can be proven that any two versions are the same, except possibly on a set of measure zero. So $$ g_1(Y(\omega)) = g_2(Y(\omega))=g_3(Y(\omega)) \mbox{ for almost all $\omega \in \Omega$} $$ where "almost all" means "except possibly on a set of measure zero."
This is similar in spirit to the fact that the exponential random variable with parameter $\lambda>0$ has an infinite number of PDF functions (nonnegative functions $f_X:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy $P[X\leq x] = \int_{-\infty}^x f_X(t)dt$ for all $x \in \mathbb{R}$), but each one of them differs from the standard one $$ f_X(x)=\left\{\begin{array}{ll} \lambda e^{-\lambda x} & \mbox{ if $x \geq 0$} \\ 0 & \mbox{ else} \end{array}\right.$$ only on a set of measure zero. None of the integration properties are changed if we change the function value at only one point, or two points, or a set of points that has measure zero. So the following function $\tilde{f}_X$ is also a valid PDF: $$ \tilde{f}_X(x) = \left\{\begin{array}{ll} \lambda e^{-\lambda x} & \mbox{ if $x \geq 0, x \neq 12.6$} \\ 42 & \mbox{ if $x=12.6$}\\ 0 & \mbox{ else} \end{array}\right.$$
So if we say "assume $E[X] = E[X|Y]$ almost surely" it means that $E[X|Y]$ is almost surely equal to the constant $E[X]$. That is, all versions of $E[X|Y]$ differ from the constant $E[X]$ only on a set of measure zero. That is, if $g(Y)$ is a version of $E[X|Y]$ then $$ P[\omega \in \Omega : g(Y(\omega)) \neq E[X]] = 0$$
Practice exercise: Define $$\Omega = \{pink, yellow, green, blue\}$$ Assume all outcomes are equally likely, so \begin{align} &P[\{pink\}]=P[\{yellow\}]= P[\{green\}] \\ &= P[\{blue\}]=1/4 \end{align} Define $X:\Omega\rightarrow\mathbb{R}$ and $Y:\Omega\rightarrow\mathbb{R}$ by \begin{align*} Y(pink) = 0 , \quad & X(pink) = -1\\ Y(yellow) = 0, \quad & X(yellow) = 1\\ Y(green) = 1, \quad & X(green) = -2\\ Y(blue) = 1, \quad & X(blue) = 2 \end{align*}
a) Compute the real number $E[X]$.
b) Compute the real numbers $E[X|Y=0]$ and $E[X|Y=1]$. Call these real numbers $m_0$ and $m_1$, respectively.
c) Are $X$ and $Y$ independent? Are they uncorrelated?
d) Note that $E[X|Y]$ can be defined as follows: Define the function $$g(y) = \left\{\begin{array}{cc} m_0 & \mbox{ if $y=0$} \\ m_1 & \mbox{ if $y=1$} \end{array}\right.$$ Then $E[X|Y]$ can be defined as $g(Y)$. Is it true that $E[X|Y](\omega)=E[X]$ for all $\omega \in \Omega$?
e) Change the problem by making $\Omega = \{pink, yellow, green, blue, tan\}$, and assume \begin{align} P[\{pink\}]&=1/4\\ P[\{yellow\}]&= 1/4\\ P[\{green\}]&=1/4\\ P[\{blue\}]&=1/4\\ P[\{tan\}]&= 0 \end{align} Define $X$ and $Y$ the same, except we make $X(tan) = Y(tan)=42$. You will find that none of the expectations are different, and now $E[X|Y]$ is "almost surely" $E[X]$, and you can define infinitely many versions of $E[X|Y]$ as $g(Y)$ via the function: $$g(y) = \left\{\begin{array}{cc} m_0 & \mbox{ if $y=0$} \\ m_1 & \mbox{ if $y=1$} \\ AnythingYouWant & \mbox{ if $y=42$} \end{array}\right.$$ You then ask "Why do we care about part (e), since probability zero outcomes don't matter"? Exactly. You should not be overly concerned about part (e). The practical meat of a problem will not change if you intuitively interpret "almost surely" as "surely." You can always change a probability space by throwing away a set of outcomes that has measure zero, which reduces the sample space but has no impact on any probability or expectation.