Showing equivalence of initial law $p_0(x,\cdot)=\delta_x$ of a transition function to $P_0f=f$

by Septime   Last Updated July 12, 2019 10:20 AM - source

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 $$P_tf(x)=\int_E \delta_{x}(dy)f(y) = \int_E \mathbb{1}_{dy}(x)f(y)= \int_{dy} f(y).$$



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