Homomorphism densities for kernels generalizes the simple graph case I am studying the well-written book Large networks and Graph Limits by László Lovász (you can find it here).
If $F$ and $G$ are two simple graphs, then their homomorphism density is defined as $t(F,G) = \dfrac{\text{hom}(F,G)}{v(G)^{v(F)}}$.
Then in section 7.2. it goes like, if $W$ is a kernel, i.e. a bounded symmetric measurable $[0;1]^2 \to \mathbb{R}$ function, $t(\cdot,W)$ is defined as if $F$ is a simple graph with $V(F)=\{1; \ldots; n\}$, then
$$ t(G,W) = \idotsint\limits_{[0;1]^n} \prod_{\{i,j\} \in E(G)} W(x_i, x_j) \, \prod_{i=1}^{n} \mathrm{d}x_i $$
Right. Then it says that it's obvious that if $F$ and $G$ are two simple graphs, then
$$ t(F,G) = t(F, W_G) $$
where $W_G$ is the kernel corresponding to the graph $G$ (visually, $W_G$ is obtained by replacing the $(i, j)$ entry in the adjacency matrix of $G$ by a rectangle of size $\frac{1}{v(G)} \times \frac{1}{v(G)}$ , and define the function value on this square as $1$).
But I just can't figure it out why... I tried to use the fact that
$$ t(F,G) = \sum_{f\colon V(F)\to V(G)} \left(\prod_{u\in V(F)} \dfrac{1}{v(G)} \prod_{\{u,v\} \in E(F)} \mathbb{I}_{\left\{\{f(u), f(v)\} \in E(G)\right\}}\right) $$
where  $\mathbb{I}_{\{\cdot\}}$ is the Iverson bracket, but I think I got nothing closer.
 A: For concreteness, suppose that $G$ has vertex set $\{0,1,\dots,N-1\}$ (starting at $0$ to make rounding easier). Then a point $(x_1, x_2, \dots, x_n) \in [0,1]^n$ corresponds to the function $f : V(F) \to V(G)$ where $f(i) = \lfloor Nx_i\rfloor$ for all $i \in V(F)$. With this correspondence, a uniform distribution over $[0,1]^n$ corresponds to a uniform distribution over functions $f : V(F) \to V(G)$.
Moreover, $W_G(x,y) = 1$ exactly when $G$ has an edge between $\lfloor Nx\rfloor$ and $\lfloor Ny\rfloor$. Therefore for a point $(x_1, x_2, \dots, x_n) \in [0,1]^n$, the product $$\prod_{ij \in E(F)} W_G(x_i, x_j)$$ is $1$ exactly when the corresponding function $f$ is a homomorphism of $G$: when for every $ij \in E(F)$, $f(i)f(j) \in E(G)$.
The integral
$$
   t(F,W_G) = \int_{[0,1]^n} \prod_{ij \in E(F)} W_G(x_i, x_j) d\mathbf x
$$
gives the probability that if we choose a random point $(x_1, x_2, \dots, x_n) \in [0,1]^n$, we'll have $ \prod_{ij \in E(F)} W_G(x_i, x_j) =1$. Under the correspondence above, this is the probability that if we choose a random $f : V(F) \to V(G)$, it will be a homomorphism.
That probability can also be written as $t(F,G) = \frac{\text{hom}(F,G)}{v(G)^{v(F)}}$ simply because there are $v(G)^{v(F)}$ possible functions, and $\text{hom}(F,G)$ of them are homomorphisms. Therefore $t(F,G) = t(F,W_G)$.
