Given a segment and a value $c$ less than the segment length, let $A\_1,\dots,A\_n$ be finite unions of intervals on the segment. We choose a finite union of intervals $B$ with $|B|=c$ that maximizes $|B\cap A\_1|\times\dots\times |B\cap A\_n|$, where $|\cdot|$ denotes the length (i.e. Lebesgue measure). If there are many such $B$, we choose one arbitrarily.


Now, we shrink $A\_1$ to $A\_1'\subseteq A\_1$, and choose $B'$ using the same procedure. Is it always true that $|B'\cap A\_1'|\le |B\cap A\_1|$?


If $A\_1,\dots,A\_n$ are **disjoint** finite unions, the answer is positive, as shown [here](https://mathoverflow.net/questions/374055/shrinking-subset-with-disjoint-unions).



Shrinking subset and product
OK, it seems that this is a counterexample.


Take 8 disjoint segments $I\_1,\dots,I\_8$ of length 1. Take 8 sets
$$
 S\_1=\{1,2,3\}, \quad
 S\_1=\{4,5,6\}, \quad
 S\_1=\{1,2,4\}, \quad
 S\_1=\{1,2,5\}, \quad
 S\_1=\{1,2,6\}, \quad
 S\_1=\{1,3,4\}, \quad
 S\_1=\{1,3,5\}, \quad
 S\_1=\{1,3,6\}.
$$
Say that $I\_i$ lies in $A\_j$ iff $j\in S\_i$, otherwise $I\_i$ and $A\_j$ are disjoint. Finally, set $c=2$.


In this situation, the optimal $B$ is $I\_1\cup I\_2$. where the product equals $1$. This follows from the fact that $\sum\_j |B\cap A\_j|\leq 6$, since any point is covered by at most three of the $A\_j$.


Now set $A\_1’=A\_1\setminus I\_1$. Consider the quantities
$$
 x=\left|B’\cap\left(\bigcup\_{i=3}^8 I\_i\right)\right|, 
 \quad
 y=|B’\cap I\_2|.
$$
Then, by AM—GM,
$$
 |B’\cap A\_1’|\leq x, \quad
 \prod\_{j=2}^3 |B’\cap A\_j|\leq (x/2+2-x-y)^2, \quad
 \prod\_{j=4}^6 |B’\cap A\_j|\leq (x/3+y)^3,
$$
and the equalities are achievable simultaneously. Hence, in the optimal case, we have
$$
 \prod\_{j=1}^6|B’\cap A\_j|=x(2-x/2-y)^2(x/3+y)^3
 =\frac{6x\cdot (36-9x-18y)^2\cdot (4x+12y)^3}{6\cdot 18^2\cdot 12^3}.
$$


So we seek for a maximizer $(x\_0,y\_0)$ of
$$
 f(x,y)= 6x\cdot (36-9x-18y)^2\cdot (4x+12y)^3
$$
under the conditions $x.y\geq 0$, $x+y\leq 2$. We claim that such a maximizer has $x\_0\geq 24/17$, which provides $|B’\cap A\_1’|>|B\cap A\_1|=1$, as desired.


Indeed, we have
$$
 f\left(\frac{24}{17},\frac{10}{17}\right)
 =\frac{144}{17}\cdot \left(\frac{216}{17}\right)^5.
$$
On the other hand, if $x\leq 24/17$, by AM—GM we have
$$
 f(x,y)\leq 6x\cdot\left(\frac{2(36-9x-18y)+3(4x+12y)}5\right)^5
 =6x\cdot\left(\frac{72-6x}5\right)^5;
$$
the right hand part is an increasing function for $0\leq x\leq 2$, so
$$
 f(x,y)\leq 6\cdot \frac{24}{17}\cdot\left(\frac{72-6\cdot 24/17}5\right)^5
 = f\left(\frac{24}{17},\frac{10}{17}\right),
$$
as desired.


