A symmetric measurable function $W\colon [0,1]^2 \to [0,1]$ is called a graphon.
Another notion is the cut norm of a symmetric measurable function $W\colon [0,1]^2 \to \mathbb{R}$: $$|| W ||_{\square} = \sup\limits_{S,T\in\mathcal{L}([0,1])} \left|\iint\limits_{S\times T} W(x,y) \ \mathrm{d}x \mathrm{d}y \right|$$ where $\mathcal{L}([0,1])$ is the set of all Lebesgue-measurable subsets of $[0,1]$.
My question is the following: if we take the cut norm of a graphon $W$, since $W\geq 0$, we have $$ ||W||_{\square} = \sup\limits_{S,T\in\mathcal{L}([0,1])} \iint\limits_{S\times T} W(x,y) \ \mathrm{d}x \mathrm{d}y $$ Furthermore, since $W$ is non-negative on the entire $[0,1]^2$, then its integral is exactly the volume under the graph of $W$, i.e. the measure of the set $\{(x,y,z) | x,y \in [0,1], 0\leq z \leq W(x,y) \} \subset \mathbb{R}^3$. And this way, I believe that the quantity $ \iint\limits_{S \times T} W(x,y) \ \mathrm{d}x \mathrm{d}y$ will be the largest when $S=T=[0,1]$, which means that $||W||_{\square} = \iint\limits_{[0,1]^2} W(x,y) \ \mathrm{d}x \mathrm{d}y $.
Am I wrong?
For a graphon, or more generally for any nonnegative $W\colon [0,1]^2 \to [0,1]$ you are correct. However, even for graphons we are not just interested in the cut norm, but also the cut metric $$ \delta_{\square}(W_1, W_2) = \inf_{\varphi} \|W_1 - W_2^{\varphi}\|_{\square} $$ where $\varphi\colon [0,1] \to [0,1]$ is a measure-preserving bijection and $W_2^{\varphi}(x,y) = W_2(\varphi(x), \varphi(y))$.
No doubt you will encounter this definition in more detail later; intuitively, $W_2^\varphi$ is a "relabeling" of $W_2$, and $\|W_1 - W_2^{\varphi}\|_{\square}$ is minimized when $W_2^{\varphi}$ is as similar to $W_1$ as possible.
However, the point for now is that even though $W_1$ and $W_2$ are both nonnegative, $W_1 - W_2^{\varphi}$ is not, and so in the computation of $\|W_1 - W_2^{\varphi}\|$ it is not necessarily best to take $S = T = [0,1]$.