# Show that $\mu\left(\bigcup_{j=1}^{N}E_j\right) \geq \frac{1}{6}\sum_{j=1}^{N}\mu(E_j)$

This came from an old qualification exam for measure theory:

Suppose $(X,M,\mu)$ is a measure space and $E_1,\ldots, E_N\in M$ with $\mu(E_j\cap E_k)\leq \mu(E_j)/N$ for each $j\neq k$. Show that $$\mu\left(\bigcup_{j=1}^{N}E_j\right) \geq \frac{1}{6}\sum_{j=1}^{N}\mu(E_j)$$

Thoughts: I thought that for $E_1,\ldots,E_n\in M$ that $$\mu\left(\bigcup_{1}^{n}E_j\right)\leq \sum_{1}^{n}\mu(E_j)$$ when $E_j$'s are not disjoint. I know we can use the disjointification trick to get equality. I am not sure how to show the latter though. Any suggestions are greatly appreciated.

Here is my try:

1) Prove by induction that: $$\mu\left(\bigcup_{j=1}^N E_i\right)\ge \sum_{j=1}^N \mu(E_j)-\sum_{i<k}\mu(E_iE_k)$$

2) From $\mu(E_i E_k)\leq \mu(E_i)/N$ we get also $\mu(E_i E_k)\leq \mu(E_k)/N$, so: $$\mu(E_iE_k)\le\frac{\mu(E_i)+\mu(E_k)}{2N}$$

3) Sum the previous inequality for $i<k$ to get: $$\sum_{i< k}\mu(E_iE_k)\le\frac{N-1}{2N}\sum_{j=1}^N \mu(E_j)$$

4) Plug in the Step 1 to get: $$\mu\left(\bigcup_{j=1}^N E_i\right)\ge \frac{N+1}{2N}\sum_{j=1}^N \mu(E_j)$$

5) Trivially prove that $\frac{N+1}{2N}\ge\frac{1}{6}$ (so it looks like you could very well have $1/2$ instead of $1/6$)

• And the factor $\frac{N+1}N$ is best possible, i.e., it can be attained for each $N.$ – bof Jan 5 '17 at 2:35
• Are these steps I need to prove? If so, I am not sure how to prove 1) – Wolfy Jan 6 '17 at 22:19
• @Wolfy Plain ol' induction. Induction step: $\mu(\bigcup_{j=1}^{n+1}E_j)=\mu(\bigcup_{j=1}^nE_j)+\mu(E_{n+1})-\mu(\bigcup_{j=1}^n(E_jE_{n+1}))\ge\sum_{i=1}^n\mu(E_j)-\sum_{1\le i<k\le n}\mu(E_iE_k)+\mu(E_{n+1})-\sum_{j=1}^n\mu(E_jE_{n+1})=\sum_{i=1}^{n+1}\mu(E_j)-\sum_{1\le i<k\le n+1}\mu(E_iE_k)$. That's all. – Momo Jan 7 '17 at 4:25