Divide a positive measure set into two sets with positive measure lying off-diagonally Let me first formulate the question that deals with $2$ parts.

Let $A\subset [0,1]^2$ be a measurable set with positive Lebesgue
measure. For every point $(a,b)\in[0,1]^2$, define
$$A_1(a,b)=\{(x,y)\in A\mid x\ge a,y\le b\}$$ and
$$A_2(a,b)=\{(x,y)\in A \mid x\le a,y\ge b\}.$$ Do there exist a point
$(a',b')\in [0,1]^2$ such that both $A_1(a',b')$ and
$A_2(a',b')$ have positive Lebesgue measure?

Note that the problem is easy if you can find an open rectangle inside $A$. But there do exist sets with positive Lebesgue measure which contains no rectangle (google Fat Cantor Set for example).
If the answer to the above question is affirmative, I would like to know if it holds for the $k$-parts problem as well.

Let $A\subset [0,1]^2$ be a measurable set with positive Lebesgue
measure. Do there exists two pairs of $k-1$-tuples $0=a_0<a_1<\cdots<a_{k-1}<a_{k}=1$ and $0=b_0<b_1<\cdots<b_{k-1}<b_{k}=1$ such that for each $\ell=1,2,\ldots,k$, the sets
$$A_{\ell}:=\{(x,y)\in A\mid a_{\ell-1} \le x\le a_{\ell},b_{k+1-\ell}\le y \le b_{k-\ell} \}$$
all have positive Lebesgue measure?

First of all I made up this problem. So, I am not sure if it is at all true. But my guess is that it is true. So, to prove it I was trying from first hand principle. Like if we denote $\lambda$ to be the Lebesgue measure on $\mathbb{R}^2$, since the function $g(r)=\lambda(A\cap ([0,1]\times [0,r]))$ is continuous, we can find $g(r_0)=\frac12\lambda(A)$. From there I got two sets of equal measure. From there I was trying similar trick. But it didn't help.
Any suggestions or help?
 A: The Lebesgue density theorem says, intuitively, that $A$ tends to fill up almost all of a sufficiently small rectangle centered at almost any point of $A$, so if it works for sets containing an open rectangle, it ought to work for all sets of positive measure.
Indeed, by the Lebesgue density theorem, for almost every $(a, b)$ of $A \cap (0,1)^2$, the density of $A$ at $(a,b)$ is $1$, i.e.
$$\lim_{\epsilon \to 0} \frac{m(A \cap ([a-\epsilon, a+\epsilon] \times [b-\epsilon, b+\epsilon]))}{4 \epsilon^2} = 1.$$
In particular, there is $\epsilon$ so small that this quotient is strictly greater than $3/4$.  By pigeonhole, this means that we must have $m(A \cap ([a, a+\epsilon] \times [b-\epsilon, b])) > 0$ and $m(A \cap ([a-\epsilon, a] \times [b, b+\epsilon])) > 0$.  Since these are subsets of $A_1, A_2$ respectively, we have the desired conclusion.
So in fact, not only does there exist $(a,b)$ as desired, but almost every $(a,b) \in A$ has this property.
I think a similar argument works for your more general statement.  Choose $\epsilon$ small enough that the quotient is greater than $1 - (k+1)^{-2}$, and arrange your points $(a_i, b_i)$ evenly on the diagonal from $(a-\epsilon, b+\epsilon)$ to $(a+\epsilon, b-\epsilon)$.  If you subdivide the square $[a-\epsilon, a+\epsilon] \times [b-\epsilon, b+\epsilon]$ into $(k+1)^2$ little squares, by pigeonhole they must all intersect $A$ with positive measure, including the $k+1$ of them on the diagonal that you care about.
