Prove that $\mu\left(\cup_{k=1}^\infty A_k\right)=\sum_{k\ge1}\mu(A_k)$ Suppose that the measurable sets $A_1,A_2,...$ are "almost disjoint" in the sense that $\mu(A_i\cap A_j) = 0$ if $i\neq j$. Prove that $$\mu\left(\cup_{k=1}^\infty A_k\right)=\sum_{k\ge1}\mu(A_k)$$
Conversely, suppose that the measurable sets $A_1,A_2,...$ satisfy $$\mu\left(\cup_{k=1}^\infty A_k\right) = \sum_{k=1}^\infty\mu(A_k)<\infty$$ Prove that the sets are almost disjoint.

Here $\mu(A)$ denotes the Lebesgue measure of $A$.
I know that if the sets $S_1,S_2,...$ are all measurable, then $$\mu(\cup_{k=1}^\infty S_k)\le\sum_{k=1}^\infty \mu(S_k)$$
and equality holds if the sets are disjoint. How can I accommodate this for almost disjoint sets?
 A: Original question: The following bounds on indicator functions hold pointwise:
$$\sum_{k\geq 1}1_{A_k}-\sum_{i\neq j}1_{A_i A_j}\leq 1_{\cup_k A_k}\leq \sum_{k\geq 1}1_{A_k}. $$ Now integrate with respect to the measure $\mu$.
Converse:  Let $X=\sum_{k\geq 1} 1_{A_k}$ and  $Y=1_{\cup_k A_k}$.
Then $X-Y\geq 0$, but the equation $\mu(\cup_k A_k)=\sum_k\mu(A_k)$
says that $\int X-Y\,d\mu=0$. Therefore $\mu(X-Y>0)=0$.
Now for $i\neq j$, we have $A_iA_j\subseteq (X-Y >0)$, so that $\mu(A_i A_j)=0$.
A: Use the fact that the measure of
$$\bigcup_{k < l} A_k \cap A_l$$
is zero.
A: We can use Bonferroni's inequalities: for $N$ and $(B_j)_{j=1}^N$ measurable sets,
$$\sum_{j=1}^N\mu(B_j)\geqslant  \mu\left(\bigcup_{j=1}^NB_j\right)\geqslant \sum_{j=1}^N\mu(B_j)-\sum_{1\leqslant i\lt j\leqslant N}\mu(B_i\cap B_j).$$
A: Define $B_1 = A_1$, $B_n = A_n - \bigcup_{k=1}^{n-1} A_k$. We have $\bigcup_{n=1}^\infty B_n = \bigcup_{n=1}^\infty A_n$ and the sets $B_n$ are disjoint. Furthermore,
$$
\mu(A_n) = \mu\left(A_n - \bigcup_{k=1}^{n-1} A_k\right) + \mu\left(A_n \cap \bigcup_{k=1}^{n-1} A_k\right).
$$
Since $\mu\left(A_n \cap \bigcup_{k=1}^{n-1} A_k\right) = 0$ by hypothesis, we have $\mu(A_n) = \mu(B_n)$. It follows that
$$
\mu\left(\bigcup_{n=1}^\infty A_n\right) =  \mu\left(\bigcup_{n=1}^\infty B_n\right) = \sum_{n=1}^\infty \mu(B_n) = \sum_{n=1}^\infty \mu(A_n).
$$

The other direction can be done similarly. Note that you'll need to rearrange an infinite series. The condition $\sum_{n=1}^\infty \mu(A_n) < \infty$ is crucial for showing that the series converges absolutely, and hence can be rearranged.
