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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?

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    If $\mu$ denotes the measure in question (e.g., the Lebesgue measure) on $\mathbb{R}^2$, then $$ \left|\int_{S \times T} F \; d\mu \right| \leq \int_{S \times T} |F| \; d\mu \leq \int_{S \times T} 1 \; d\mu = \mu(S \times T). $$ This gives you an idea of what measure $S \times T$ should have for your statement to be true. – 2017-02-16
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    measure of $S \times T$ can be any number in $[0, 1]$ in the statement I want to prove. – 2017-02-16
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    In that case, your statement may not hold. For example, with Lebesgue measure, take $S \times T = [0, 1]^2$, $F = 1$ constantly, and $\epsilon = 0.1$. Answerer tofurind, below, provides another counterexample. – 2017-02-16
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    I might have mislead with the probelm statement, i rephrased it. We do know, that $\vert \int_{S \times S} F \vert \leq \varepsilon$ for all measureable S, and we are trying to show that the result follows from that. – 2017-02-16
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    Then you'd have to have, in particular, the bound, $$ \left| \int_{[0, 1]^2} F \right| \leq \epsilon. $$ Also, I would take a look at what the positive and negative parts of $F$ would have to be like for your $\leq \epsilon$ premise to hold. Is the condition $|F| \leq \epsilon$ too strong? – 2017-02-16
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    Its a sufficient, but I believe, not a necessary condition. In that case $\vert \int_{S \times T} F \vert $ would be $\leq \varepsilon$ too. I see this statement as: every square $S \times S$ sums to something less in absolute value than $\varepsilon$. When we extend this to any rectangle, not just diagonal square, we lose some of the bound. But not much. Like some sort of triangle inequality, but i failed to apply them here. – 2017-02-16

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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*}

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    That's a good observation. I'm working with graphons, so the functions have to by symmetric. I corrected the question. I found this statement in paper by Lovasz, and can't put the details together for this implication. – 2017-02-16
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    Try something like this (extracting the "symmetric part" from $S \times T$): $$ \left|\int_{S \times T} F\right| = \left| \int_{(S \cap T) \times (S \cap T)} F + \int_{(S \times T) \setminus ((S \cap T) \times (S \cap T))} F \right| \leq \left| \int_{(S \cap T) \times (S \cap T)} F\right| + \left|\int_{(S \times T) \setminus ((S \cap T) \times (S \cap T))} F \right| $$ – 2017-02-16
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    I have added a proof for a weaker estimate. Maybe this is sufficient for your purposes? – 2017-02-17