Given a segment and a value $c$ less than the segment length, let $A\_1,\dots,A\_n$ be **disjoint** 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 $|B'\cap A\_1'|\le |B\cap A\_1|$?



Shrinking subset with disjoint unions
$\newcommand\ta{\tilde a}
\newcommand\tb{\tilde b}
\newcommand{\ep}{\varepsilon}$The answer is yes at least in the case when the $A\_i$'s are pairwise disjoint. Indeed, then we can restate the problem as follows (with $a\_i$ in place of $|A\_i|$ and $b\_i$ in place of $|B\cap A\_i|$):



> 
> Take any $a=(a\_1,\dots,a\_n)\in(0,\infty)^n$. Let $c\in(0,\sum\_i a\_i)$. Let
> $$B(a):=B\_c(a):=\{b=(b\_1,\dots,b\_n)\colon \sum\_i b\_i=c, 0\le b\_j\le a\_j\ \forall i\}.$$
> The set $B\_c(a)$ is compact and nonempty, since $ta\in B\_c(a)$ for $t:=c/\sum\_i a\_i\in(0,1)$.
> 
> 
> 



> 
> Let then $b=(b\_1,\dots,b\_n)$ be any maximizer of $\pi(b):=b\_1\cdots b\_n$ over all $b\in B(a)$. In view of the last sentence of the previous paragraph, such a maximizer $b$ exists and $\pi(b)\in(0,\infty)$ for such $b$.
> 
> 
> 



> 
> Take any $j\in[n]:=\{1,\dots,n\}$ and replace $a\_j$ by some real
> \begin{equation}
> \ta\_j\ge a\_j, \tag{1} 
> \end{equation}
> to get $\ta:=(\ta\_1,\dots,\ta\_n)$, where $\ta\_i:=a\_i$ for $i\ne j$. Let $\tb=(\tb\_1,\dots,\tb\_n)$ be any maximizer of $\pi(b)$ over all $b\in B(\ta)$. Does it then follow that $\tb\_j\ge b\_j$?
> 
> 
> 


Let us show that the answer is yes. Indeed, without loss of generality (wlog)
$$a\_1\ge\dots\ge a\_n>0.$$


If $b\_i<b\_{i+1}$ for some $i\in[n-1]$, then $0\le b\_i<b\_{i+1}\le a\_{i+1}\le a\_i$, and so,
we can replace $b\_i,b\_{i+1}$ by $b\_i+\ep,b\_{i+1}-\ep$ with $\ep\in(0,\min(a\_i-b\_i,\frac{b\_{i+1}-b\_i}2))$, thus still satisfying the conditions on $b$ while making the value of $\pi$ greater, which contradicts the assumption that $b$ is a maximizer of $\pi$. So,
\begin{equation}
 b\_1\ge\dots\ge b\_n>0.\tag{1.5}
\end{equation}


Since $0\le b\_j\le a\_j$ for all $j$ and $\sum\_i b\_i=c<\sum\_i a\_i$, there is some $k\in[n]$ with $b\_k<a\_k$. If $k\ge2$ and $b\_{k-1}>b\_k$, then we can replace $b\_{k-1},b\_k$ by $b\_{k-1}-\ep,b\_k+\ep$
with $\ep\in(0,\min(a\_k-b\_k,\frac{b\_{k-1}-b\_k}2))$, thus still satisfying the conditions on $b$ while making the value of $\pi$ greater, which contradicts the assumption that $b$ is a maximizer of $\pi$. So,
$$b\_k<a\_k\implies b\_{k-1}=b\_k<a\_k\le a\_{k-1}\implies b\_{k-1}<a\_{k-1}\implies\cdots.$$
So, there is some $m\in[n]$ such that
\begin{equation}
 b\_1=\cdots=b\_m>b\_{m+1}=a\_{m+1}\ge\cdots \ge b\_n=a\_n>0,\tag{2}
\end{equation}
whence
\begin{equation}
 b\_i=\min(b\_1,a\_i)\tag{3}
\end{equation}
for all $i$.


Let us say that $i\in[n]$ is a drop point for $a$ if either (i) $i=1$ or (ii) $i\ge2$ and $a\_{i-1}>a\_i$. Let us then define the drop value at a drop point $i$ as $a\_{i-1}-a\_i$ if $i\ge2$ and as $\infty$ if $i=1$. For each $j\in[n]$, there is always a permutation of the indices leaving $a$ invariant after which $j$ becomes a drop point (of $a$). So, wlog $j$ in (1) is a drop point. Moreover, we may assume that $\ta\_j-a\_j$ is no greater than the drop value at $j$ (otherwise, increasing $\ta\_j$ continuously until $\ta\_j-a\_j$ reaches the drop value at $j$, we switch from $j$ to the next drop point -- say $j\_1$ -- of $a$ to the left of $j$ and continue continuously increasing $\ta\_{j\_1}$, etc.). Thus, wlog
$$\ta\_1\ge\dots\ge\ta\_n>0.$$


So, similarly to (1.5), we get $\tb\_1\ge\dots\ge \tb\_n>0$ and hence,
similarly to (3), we get
\begin{equation}
 \tb\_i=\min(\tb\_1,\ta\_i)\tag{4}
\end{equation}
for all $i$.


Take any $j\in[n]$ such that (1) holds. We have to show that then $\tb\_j\ge b\_j$. Suppose the contrary:
\begin{equation}
\tb\_j<b\_j. \tag{5} 
\end{equation}
Then $\tb\_j<b\_j\le a\_j\le\ta\_j$.


Suppose now that $\exists i\in[n]\ \tb\_i>\tb\_j$. Then we can replace $\tb\_i,\tb\_j$ by $\tb\_i-\ep,\tb\_j+\ep$ with $\ep\in(0,\min(\ta\_j-\tb\_j,\frac{\tb\_i-\tb\_j}2))$, thus still satisfying the conditions on $\tb$ while making the value of $\pi$ greater, which contradicts the assumption that $\tb$ is a maximizer of $\pi$.


So, $\tb\_j\ge\tb\_i$ for all $i$. So, in view of (4), (5), and (3), $\tb\_1=\tb\_j<b\_j\le b\_1$, whence, by
(3) and (4) $\tb\_i\le b\_i$ for all $i$, with $\tb\_1<b\_1$, so that $c=\sum\_i\tb\_i<\sum\_i b\_i=c$.


Thus, (5) leads to a contradiction. $\Box$.


