# 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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