Prove $\sum_{i,j:iThe problem is:

Assume that $(X,\mathcal{A} , μ)$ is a finite measure space (i.e., $μ(X) < \infty $ ) and the sequence of sets $E_{j}\in \mathcal{A}$ satisfies $\sum^\infty_{j=1} μ(E_{j} )=\infty$. Show that then $\sum_{i,j:i<j} μ(E_{i} \cap E_{j} )=\infty$.

I tried to assume that $\sum_{i,j:i<j} μ(E_{i} \cap E_{j} )<\infty$ to get a contradiction, and I know that we can subtract since the measure is finite. I need help! I have no idea how to solve it.
Thanks in advance.
 A: Hint: Let $\chi_i$ be the indicator function of $E_i$, and $f_N = \sum_{i=1}^N \chi_i$.  When $f_N = m$, $\sum_{i=1}^N \sum_{j=i+1}^N \chi_i \chi_j = m(m-1)/2$.
A: Just to complement what Robert Israel said
$$f^2_N=\big(\sum^N_{j=1}\mathbb{1}_{E_j}\Big)^2=\sum^N_{j=1}\mathbb{1}_{E_j}+2\sum^N_{1\leq i <j\leq N}\mathbb{1}_{E_j}\mathbb{1}_{E_i}=f_N+2\sum^N_{1\leq i <j\leq N}\mathbb{1}_{E_j}\mathbb{1}_{E_i}$$
whence
$$\sum_{1\leq i <j\leq N}\mathbb{1}_{E_i\cap E_j}=\frac{f_N(f_N-1)}{2}\geq \frac12f_N\mathbb{1}_{\{f_N>1\}}=\frac12f_N-\frac{1}{2}\mathbb{1}_{\{f_N=0\}\cup\{f_N=1\}}$$
Integrating yields
$$\begin{align}\sum_{1\leq i <j\leq N}\mu(E_i \cap E_j)&\geq\frac12\int f_N\,d\mu-\frac{1}{2}\mu(f_N\in\{0,1\})\\
&\geq\frac{1}{2}\sum^N_{i=1}\mu(E_j)-\frac12\mu(X)\xrightarrow{N\rightarrow\infty}\infty
\end{align}$$
A: Here's an answer of a slightly different flavor, with no indicator functions involved. The idea here will be that since $X$ has finite measure, then the parts where the $E_i$'s do not overlap must be finite in measure as well. It will follow that the sum of overlaps is infinite.
Let's fix an $E_i$. Define $C_i = \bigcap_{i\neq j} E_j^c$, the complement of all the $E_j$'s other than $E_i$. Then we can write$$
E_i = \left(\bigcup_{j\neq i}E_i \cap E_j\right)\cup (E_i \cap C_i).
$$ From this we find that$$
\mu(E_i) \leq \sum_{j\neq i}\mu(E_i \cap E_j) + \mu(E_i \cap C_i).
$$
Now sum both sides over all $i$ and compare: $$
\infty = \sum_i \mu(E_i) \leq \sum_i\sum_{j\neq i}\mu(E_i \cap E_j) + \sum_i \mu(E_i \cap C_i) $$
$$
= 2\sum_{\substack{i,j\\ i<j}}\mu(E_i \cap E_j) + \sum_i \mu(E_i \cap C_i)
$$
But the $E_i \cap C_i$ are disjoint (since $E_i \cap C_i \subseteq E_i$ and $E_j \cap C_j \subseteq E_i^c$ for $i\neq j$). So since $\mu(X)$ is finite, so is $\sum_i \mu(E_i \cap C_i)$. It follows that the sum in question is infinite.
