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

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.

• "I think we need to assume that the intersection is nonempty." It seems like this is something you'll be proving, not assuming (the conclusion of course implies this). Is it possible for the intersection to be empty given the premises? Aug 18, 2017 at 16:12
• I am a bit confused---what is the measure of a sum? Do you mean the measure of the union, or the sum of the measures of the sets? Aug 18, 2017 at 16:18
• Corrected the typo. Aug 18, 2017 at 16:24
• Your $n=2$ case is incorrect. What you have is $m(E_1)+m(E_2) > 2 -1 = 1$... Aug 18, 2017 at 17:10
• I am not clear on why you didn't accept Marios Gretsas' proof. Sure, it is by contradiction, but making it direct is a simple exercise. Aug 21, 2017 at 1:16

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 $$\sum_{k=1}^nm(E_k) \leq n-1$$

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, $$m\left( \bigcap_{k=1}^n E_k \right) > 0 .$$

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

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