Symmetric measurable $S\subseteq[0,1]^2$ with $\lambda(S_x)=x$ for almost all $x\in[0,1]$

Question

Let $S\subseteq [0,1]^2$ be a measurable set that satisfies the following.

• It is symmetric, i.e., $(x,y)\in S$ if and only if $(y,x)\in S$ for every $(x,y)\in [0,1]^2$.
• For almost all $x\in[0,1]$, the ($1$-dimensional Lebesgue) measure of the $x$-section of $S$, i.e., the set $S_x=\{y\in [0,1]:(x,y)\in S\}$, is equal to $x$. (Note that since $S$ is measurable, $S_x$ is measurable for almost all $x\in[0,1]$.)
  

Does it follow that for almost all $(x,y)\in[0,1]^2$, we have $(x,y)\in S$ if and only if $x+y\geq 1$? In other words, is $S$ equal to $\{(x,y)\in[0,1]^2:x+y\geq 1\}$ up to a null-set?

Answer 1

You can reduce to a finitary problem similar to John Dawkins' comment.

Lemma. If $A_{i,j}\in [0,1]^{n\times n}$ satisfies $$\sum_{i=1}^n A_{i,j}=\sum_{i=1}^n A_{j,i}=j-\tfrac 1 2\text{ for all $1\leq j\leq n$.}$$ then $$A_{i,j}=\begin{cases} 0&\text{ when }(i-\tfrac 1 2)+(j-\tfrac 1 2)<n\\ \tfrac 1 2&\text{ when }(i-\tfrac 1 2)+(j-\tfrac 1 2)=n\\ 1 &\text{ when }(i-\tfrac 1 2)+(j-\tfrac 1 2)>n. \end{cases}$$ Proof. The first column sums to $\tfrac 1 2$, so $A_{1,n}\leq \tfrac 1 2$, but the last row sums to $n-\tfrac 1 2$, so $A_{1,n}\geq \tfrac 1 2$. These inequalities match so they become equalities, which shows that $A_{i,j}$ takes the correct values when $j=n$, and a symmetric argument shows that the values are also correct when $i=n$. The result follows by induction on $n$ and removing the last row and column.

To get the measure theoretic result, use Fubini's theorem to show that for each $n$ the matrix defined by $A_{i,j}=n^2 \mu(S\cap ([\tfrac {i-1}n,\tfrac i n]\times[\tfrac {j-1}n,\tfrac j n]))$ satisfies the conditions of the lemma. This gives $\mu(S\cap Q)=\mu(S^*\cap Q)$ for all axis-aligned squares $Q$ with rational co-ordinates, where $S^*=\{(x,y)\in[0,1]^2:x+y\geq 1\}$.

Such squares generate the sigma algebra so this shows that the indicator functions $\chi_S$ and $\chi_{S^*}$ are equal almost everywhere, which implies that $S\triangle S^*$ is a null set.

It is not necessary to assume that $S$ is symmetric, just that the "row sums" and "column sums" are correct almost everywhere.

Answer 2

Here is a proof without any discretization. Let $\chi_S$ be the indicator function of $S$. Consider $$I(S)=\int_{[0,1]^2} (2\chi_S(x,y)-1)(x+y-1).$$

We can show that $I(S)=I(S^*)$ where $S^*=\{(x,y)\in[0,1]^2:x+y\geq 1\}$: $$ \begin{align*} I(S)&=\int_0^1\int_0^1 (2\chi_S(x,y)-1)(x-\tfrac 1 2) dydx+\int_0^1\int_0^1 (2\chi_S(x,y)-1)(y-\tfrac 1 2) dxdy\\ &=2 \int_0^1\int_0^1 (2\chi_S(x,y)-1)(x-\tfrac 1 2) dydx\text{ (by symmetry of $S$)}\\ &=2 \int_0^1(2\int_0^1 \chi_S(x,y)dy-1)(x-\tfrac 1 2) dx\\ &=I(S^*) \end{align*} $$ using$\int_0^1 \chi_S(x,y)dy=\int_0^1 \chi_{S^*}(x,y)dy$ for almost every $x$. (In fact $I(S^*)=1/3$, but that is irrelevant to the proof.)

On the other hand, $$|(2\chi_S-1)(x+y-1)|=|x+y-1|=(2\chi_{S^*}-1)(x+y-1)$$ so, taking absolute values: $$I(S)\leq \int_{[0,1]^2} (2\chi_{S^*}-1)(x+y-1) = I(S^*).$$

But we know $I(S)=I(S^*)$, so $$(2\chi_S(x,y)-1)(x+y-1)=(2\chi_{S^*}-1)(x+y-1)$$ almost everywhere. Since $x+y\neq 1$ almost everywhere, this forces $\chi_S(x,y)=\chi_{S^*}(x,y)$ almost everywhere, which implies that $S\triangle S^*$ is a null set.

Again the symmetry isn't really necessary - the proof could be modified to only use that the $x$ and $y$ marginals are correct.

Answer 3

There is a simple proof that uses more directly the given information:

For $0<t<1$, consider the rectangles $A = [0,t] \times [0,1]$ and $B=[0,1]\times [1-t,1]$. Then by the given information on sections of $S$ we have using Fubini: $\lambda(S\cap A) = \int_0^t x\;dx = \frac{t^2}{2}$ and $\lambda (S\cap B) = \int_{1-t}^2 y\; dy = t-\frac{t^2}{2}$ so: $$ \lambda(S\cap A\cap B)\leq \lambda (S\cap A) \leq \frac{t^2}{2} .$$ On the other hand $\lambda(B\setminus A)= \lambda([t,1]\times[1-t,1]) = t-t^2$ so $$\lambda(S\cap A\cap B) = \lambda(S\cap B) - \lambda ( S\cap (B\setminus A)) \geq \lambda(S\cap B) - \lambda(B\setminus A) = \frac{t^2}{2}.$$ We must therefore have equality everywhere so in particular: $$ \lambda(S\cap (B\setminus A)) = \lambda(B\setminus A)\; \mbox{ and } \; \lambda (S\cap (A\setminus B)) = 0.$$ In other words for any $t\in (0,1)$, $S$ has full measure when restricted to the set $B\setminus A= [t,1]\times [1-t,1]$ and zero measure in $A\setminus B = [0,t]\times [0,1-t]$. Varying $t$ we see that a.e. point in $\{(x,y) \in [0,1]\times[0,1] : x+y\geq 1 \}$ belongs to $S$ (and only a zero measure subset in the complement).