Dirichlet forms So I'm seeing a Dirichlet form written as
$$\mathscr{E}(f,f) = \frac{1}{2} \sum_{x,y} |f(x)-f(y)|^2 K(x,y)\pi(x)$$
where $K(x,y)$ is the probability of taking a step to state $y$ from $x$.
And the conventional way of writing it seems to be
$$\mathscr{E}(f) = \int \left|\triangledown f\right|^2 d\mu. $$
How come the two expressions are equivalent.  I just don't see how $K(x,y)\pi(x)$ could be analogous to $d\mu$ like $(f(x)-f(y))^2$ is analogous to $|\triangledown f|^2.$  If it were $\pi$, I could understand, but why are they using $K(x,y)\pi(x)?$
I hope this isn't a stupid question (I feel like it is).
 A: A short introduction was given last year at PIMS Summer School in Probability by Zhenqing Chen talking about Dirichlet Form Theory and Invariance Principle, see the slides of the first lecture.
A translation from the continuous setting you know to the discrete setting is as follows. Consider a graph $G=(V(G),E(G))$ with vertex set $V(G)$ and edge set $E(G)$ and a Markov chain on $G$ of kernel $K$. This means that $K$ is defined on $V(G)\times V(G)$ and that $K(x,y)=0$ if $(x,y)\notin E(G)$. The gradient of a function $f$ defined on $V(G)$ is the function $\nabla f$ defined on $E(G)$ by
$$
\nabla f(x,y)=f(y)-f(x).
$$
Then $\mathcal{E}(f)$ is simply the $\ell^2$ norm of $\nabla f$ with respect to the measure $\mu$ defined on $E(G)$ by $\mu(x,y)=\pi(x)K(x,y)$. 
In the continuous case $\mathcal{E}(f)$ for a function $f$ defined on $\mathbb{R}^n$ (which is the analogue of $V(G)$ in the discrete setting) is the $L^2$ norm of the usual gradient $\nabla f$, defined on the tangent space $T_x\mathbb{R}^n$ (which is the analogue of $E(G)$), but the difference is not visible because at every $x$ in $\mathbb{R}^n$ the tangent space $T_x\mathbb{R}^n$ is $\mathbb{R}^n$ itself.
