I have this statement I'm trying to prove for a while:
Let $F : [0, 1]^2 \mapsto [-1, 1]$ [EDIT: be as symmetric measureable function $F(x, y) = F(y, x)$.] Im trying to show that:
If $$ \left\vert \int_{S \times S} F \right\vert \leq \varepsilon$$
,for all measureable $ S \subseteq [0, 1]$,
then,
$$ \left\vert \int_{S \times T} F \right\vert \leq 2 \varepsilon,$$
for all measureable, not neccesarily disjoint subsets $S, T \subseteq [0, 1]$.
Any ideas?
Maybe its not true for arbitrary $\varepsilon>0$ and the Lebesgue measure. Try $F=\chi_A-\chi_B$ (with characteristic function) with $A:=\{x\in[0,1]^2:x_2\geq x_1\}$ and $B:=\{x\in[0,1]^2:x_2\leq x_1\}$. Because of symmetry reasons you should get $\left\vert \int_{S \times S} F \right\vert =0$ for all measurable $S$ in $[0,1]$. But for $S=[\frac{1}{2},1]$ and $T=[0,\frac{1}{2}]$ you get $\left\vert \int_{S \times T} F \right\vert = \frac{1}{4}$.
UPDATE: I think I have a proof now for symmetric $F$ and the weaker estimate $\left\vert \int_{S \times T} F \right\vert \leq \frac{11}{2} \varepsilon$.
$\textbf{Proof.}$ Let $F:[0,1]^2\rightarrow[0,1]$ symmetric, measurable with $|\int_{M\times M}F|<\varepsilon$ for some $\varepsilon>0$ and all measurable $M\subset[0,1]$.
First step: Let $S$ and $T$ be disjoint and measurable subsets of $[0,1]$, then we obtain $$(S\uplus T)\times(S\uplus T) = S\times T \uplus T\times S \uplus S\times S \uplus T\times T.$$ Because $F$ is symmetric and with the disjointness we obtain \begin{align*} \left|\int_{(S\uplus T)\times(S\uplus T)\setminus (S\times S \uplus T\times T)} F\right| &=\left|\int_{S\times T \uplus T\times S} F \right|\\ &=\left|\int_{S\times T} F + \int_{T\times S} F \right|\\ &=2\left|\int_{S\times T} F\right|. \end{align*} Again with disjointness, the triangle inequality (and the assumptions) this leads to \begin{align*} \left|\int_{S\times T} F\right| &= \frac{1}{2} \left|\int_{(S\uplus T)\times(S\uplus T)\setminus (S\times S \uplus T\times T)} F\right| \\ &= \frac{1}{2} \left|\int_{(S\uplus T)\times(S\uplus T)}F - \int_{S\times S}F - \int_{T\times T}F\right| \\ &\leq \frac{1}{2} \left( \left|\int_{(S\uplus T)\times(S\uplus T)}F\right| + \left|\int_{S\times S}F\right| + \left|\int_{T\times T}F\right|\right) \\ &< \frac{1}{2} \left( \varepsilon + \varepsilon + \varepsilon \right) = \frac{3}{2}\varepsilon. \end{align*} As a conclusion, we obtain $\left|\int_{S\times T} F\right|<\frac{3}{2}\varepsilon$ for a disjoint pair $S$ and $T$. Now we allow $S$ and $T$ not to be disjoint. Define $A:=S\setminus T$, $B:=T\setminus S$, $C=S\cap T$ then we have \begin{align*} &S = (S\setminus T)\uplus (S\cap T)=A\uplus C,\\ &T = (T\setminus S)\uplus (S\cap T)=B\uplus C. \end{align*} This leads to \begin{align*} S\times T &= (A\uplus C) \times (B\uplus C)\\ &= (A\times B)\uplus (C\times C)\uplus (C\times B)\uplus (A\times C). \end{align*} We observe that $$A\cap B=C\cap B=A\cap C=\emptyset$$ and thus we finally have with the first step (resp. the assumptions) \begin{align*} \left|\int_{S\times T} F\right| &= \left|\int_{A\times B} F + \int_{C\times C} F + \int_{C\times B} F + \int_{A\times C} F\right|\\ &\leq \left|\int_{A\times B} F\right| + \left|\int_{C\times C} F \right|+ \left|\int_{C\times B} F\right| + \left|\int_{A\times C} F\right|\\ &\leq \frac{3}{2}\varepsilon + \varepsilon + \frac{3}{2}\varepsilon + \frac{3}{2}\varepsilon\\ &= \frac{11}{2}\varepsilon. \end{align*}