Prove that $\mu\left(\cup_{k=1}^\infty A_k\right)=\sum_{k\ge1}\mu(A_k)$

Question

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.

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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?

Answer 1

Use the fact that the measure of $$\bigcup_{k < l} A_k \cap A_l$$ is zero.

Answer 2

We can use Bonferroni's inequalities: for $N$ and $(B_j)_{j=1}^N$ measurable sets, $$\sum_{j=1}^N\mu(B_j)\geqslant \mu\left(\bigcup_{j=1}^NB_j\right)\geqslant \sum_{j=1}^N\mu(B_j)-\sum_{1\leqslant i\lt j\leqslant N}\mu(B_i\cap B_j).$$

Answer 3

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 $$ \mu\left(\bigcup_{n=1}^\infty A_n\right) = \mu\left(\bigcup_{n=1}^\infty B_n\right) = \sum_{n=1}^\infty \mu(B_n) = \sum_{n=1}^\infty \mu(A_n). $$

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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.

Answer 4

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 $\mu(A_i A_j)=0$.