Statement $1$: Let $E_1,...,E_n$ be measurable subsets of $[0,1]$ such that $$m\left( \sum_{k=1}^n E_k \right) >n-1.$$ is it true that $m\left( \bigcap_{k=1}^n E_k \right)$ has positive measure?
The answer to the above statement is no, as $$E_1=E_2 = C$$ where $C$ is the Cantor Middle-Third set in $[0,1]$ serves as a counterexample. Indeed, $$E_1+E_2 = C+C = [0,2]$$ has measure $2>1$ but $C$ has measure $0.$
However, by imposing condition that each $E_k$ has positive measure, will the modified statement hold. In particular,
Question: Let $E_1,...,E_n$ be measurable subsets of $[0,1]$ such that $$m\left( \sum_{k=1}^n E_k \right)>n-1 \quad \text{and} \quad m(E_k)>0 \text{ for each } 1\leq k\leq n.$$ Is it true that $m\left( \bigcap_{k=1}^n E_k \right)$ has positive measure?
For the question, I could not come up with a counterexample to disprove it as the cantor set does not work here.
However, I also could not provide a proof to show the question is true.
Any hint is appreciated.
For $1\le k\le n$, let $$ A_k=k\sqrt 2+\Bbb Q,$$ let $$A=\bigcup_k A_k $$ and finally $$E_k=\left(\left[\tfrac {k-1}n,\tfrac {k}n\right)\setminus A\right)\;\cup\; \bigl([0,1]\cap A_k \bigr)$$ Then the $E_k$ have measure $\frac 1n$, are even pairwise disjoint, but $\sum E_k$ differs from $[0,n]$ by only countably many points.