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

Question

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.

Answer 1

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

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

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 $1/6$)