Inequality for integration over unit square 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?
 A: 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*}
