If $\ \sum_{k=1}^n m(E_k) > n-1,$ then prove that $\bigcap_{k=1}^n E_k$ has positive measure.

Question

Question: Let $E_1,E_2,...,E_n$ be measurable subsets of $[0,1]$ with $$ \sum_{k=1}^n m(E_k) > n-1.$$ Prove that $\bigcap_{k=1}^n E_k$ has positive measure.

This is one of the questions in graduate analysis past year paper. I think we need to assume that the intersection $\bigcap_{k=1}^n E_k$ is nonempty. Otherwise, the question is false.

Anyway, let's assume that $\bigcap_{k=1}^n E_k \neq \emptyset.$

If $n=2,$ by inclusion-exclusion principle and assumption, $$m(E_1\cup E_2) + m(E_1 \cap E_2) = m(E_1) + m(E_2) > 2.$$ Since $E_1,E_2 \subseteq [0,1],$ by monotonicity of Lebesgue measure, $$m(E_1) \leq 1, m(E_2) \leq 1.$$ By finite subadditivity of Lebesgue measure, $$m(E_1 \cup E_2) \leq m(E_1) + m(E_2) \leq 2.$$

Hence, $$m(E_1 \cap E_2) > 2 - m(E_1 \cup E_2) \geq 0$$

Therefore, the intersection $\bigcap_{k=1}^n E_k$ has positive measure for $n=2.$

I try prove the general $n$ using induction, but it seems a bit long.

Does there exist an efficient method to solve the question?

EDIT: Actually I am looking for a direct proof instead of indirect proof or inductive proof. If a direct proof is not possible, then I will accept Marios's answer.

Answer 1

Let $A=\bigcap_{k=1}^nE_k$

Assume that $m(A)=m(\bigcap_{k=1}^nE_k) =0 $.

Then $1=m([0,1]$ \ $A) \leq \sum_{k=1}^nm([0,1]$ \ $E_k) =n -\sum_{k=1}^nm(E_k) $

From this we see that $$\sum_{k=1}^nm(E_k) \leq n-1$$

which is a contradiction.

Answer 2

The following is a direct proof, credited to Marios.

Observe that $$m\left([0,1] \setminus \bigcap_{k=1}^n E_k \right) \leq \sum_{k=1}^nm([0,1] \setminus E_k) = n - \sum_{k=1}^n m(E_k) < n - (n-1) = 1.$$ Therefore, $$m\left( \bigcap_{k=1}^n E_k \right) > 0 .$$

Answer 3

Basically repeating the ideas below in one line: $$m(\cap_{j=1}^n E_j)=1- m(\cup_{j=1}^n E_j^c) \ge 1-\sum_{j=1}^n m(E_j^c)=1-n +\sum_{j=1}^n m(E_j)>1-n+n-1=0.$$

Answer 4

You wrote:

"If n=2, by inclusion-exclusion principle and assumption, m(E1∪E2)+m(E1∩E2)=m(E1)+m(E2)>2"

m(E1) <= 1, m(E2) <= 1 so m(E1) + m(E2) <= 2

The conjecture is trivial for n = 1, so the assumption in the induction step for n = 2 would be

m(E1∪E2)+m(E1∩E2)=m(E1)+m(E2)>1 (= n-1 when n is 2)

Hint: for your induction step: If sum(m(Ek)) for k from 1 to n is >n−1, then the sum(m(Ek)) from 1 to n-1 must be greater than n-2, so the intersection of the sets from E1 to En-1 has positive measure.

Then use the same type of analysis as going from 1 to 2