I am reading Lovasz's book "Large networks and graph limits", and encountered the exercise that the stepping operator for graphons is contractive under the cut norm:
$$||W_P||_\square\leq||W||_\square.$$
Originally, the stepping operator from some partition $P:[0,1]=X_1\cup\cdots\cup X_k$ acting on some graphon $W:[0,1]^2\rightarrow\mathbb{R}$, which is bounded, measurable and symmetric, is defined as below:
$$W_P(x,y)=\frac{1}{\lambda(X_i)\lambda(X_j)}\int_{X_i\times X_j}W(x,y)dxdy,\forall (x,y)\in X_i\times X_j,$$
where $\lambda$ is the Lebesgue measure. And the cut norm of a graphon is defined as
$$||W||_\square=\mathop{\sup}\limits_{S,T\subset [0,1]}|\int_{S\times T}W(x,y)dxdy|,$$
where $S,T$ are both measurable. I have figured out how to do this, starting by first prove that the supremum in the cut norm of $W_P$ is the same as the supremum on all $S,T$ generated by the partition. This is done by considering each small pieces of maximum $S,T$(merely measurable) in each partition parts $X_i$, moving them using measure-preserving maps to form intervals(this can be done since when we work with $W_P$, it's constant on every small blocks), and discussing the boundaries.
Now I am thinking that, since the stepping operation can be viewed as taking conditional expectation given the $\sigma$-algebra generated by $\mathcal{P}$:
$$W_P=\mathbb{E}[W\mid\mathcal{P}\otimes\mathcal{P}],$$
here I am using $\mathcal{P}$ for the $\sigma$-algebra generated by $P$. Does this general statement below also hold?
Fix some probability triple $(\Omega,\mathcal{F},\mathbb{P})$, and a bounded, real-valued (symmetric)random variable $X$ on the product probability space $(\Omega,\mathcal{F},\mathbb{P})\otimes(\Omega,\mathcal{F},\mathbb{P})$. Let $\mathcal{G}$ be some sub-$\sigma$-algebra. Do we have the following: $$\mathop{\sup}\limits_{S,T\in\mathcal{F}}|\mathbb{E}[\mathbb{E}(X\mid\mathcal{G}\otimes\mathcal{G})1_{S\times T}]|=\mathop{\sup}\limits_{S',T'\in\mathcal{G}}|\mathbb{E}[\mathbb{E}(X\mid\mathcal{G}\otimes\mathcal{G})1_{S'\times T'}]|?$$
I tried to mimick the proof for graphons, which means $\mathcal{G}$ is generated by finite number of sets. So at first I have
$$\mathbb{E}[\mathbb{E}(X\mid\mathcal{G}\otimes\mathcal{G})1_{S\times T}]=\mathbb{E}[X\mathbb{E}(1_{S\times T}\mid\mathcal{G}\otimes\mathcal{G})],$$
and $\mathbb{E}(1_{S\times T}\mid\mathcal{G}\otimes\mathcal{G})=\mathbb{E}(1_S\mid\mathcal{G}\otimes\mathcal{G})\otimes\mathbb{E}(1_{T}\mid\mathcal{G}\otimes\mathcal{G})$, since they equal on cylinder sets in $\mathcal{G}\otimes\mathcal{G}$. THen I got stuck, since I don't know the structure of \mathbb{E}(1_S\mid\mathcal{G}\otimes\mathcal{G}), unlike in finite sub-$\sigma$-algebra case where I can discuss one-by-one in blocks.
Any help is desired.
