Let $(p_t)_{t\ge 0}$ be probability kernels und let $P_t$ be the transition operator defined by $$P_tf(x):= \int p_t(x,dy)f(y),\quad x\in E,f\in B(E).$$
I want to show the equivalence of $p_0(x,\cdot)=\delta_x$ and $P_0f=f$.
My problem is that I don't understand the meaning of $p_t(x,dy)$. Why is $dy$ inside the function? How would I integrate if I had a numerical example because $dy$ should be a set in this case. I tried the case $p_0(x,\cdot)=\delta_x$. Then we get $$P_tf(x)=\int_E \delta_{x}(dy)f(y) = \int_E \mathbb{1}_{dy}(x)f(y)= \int_{dy} f(y).$$
$A \to p_t(x,A)$ is a measure for fixed $t$ and $x$. $\int p_t(x,dy)f(y)$ is the notation for the integral of $f$ w.r.t this measure. When $t=0$ this measure is nothing but $\delta_x$ so $\int p_t(x,dy)f(y)$ is the integral of $f$ w.r.t. $\delta_x$. The value of this integral is $f(x)$.