If $\lambda(A_n \cap A_k) = 0$ then $\lambda \left( \bigcup_{n=0}^{\infty} A_n \right) = \sum_{n=0}^{\infty} \lambda(A_n)$

Question

Let $A_n$ be borel set such that $\lambda(A_n \cap A_k) = 0 \quad \mbox{for} \quad n\neq k$. $\lambda$ is Lebesgue measure. Show that $$\lambda \left( \bigcup_{n=0}^{\infty} A_n \right) = \sum_{n=0}^{\infty} \lambda(A_n)$$

I would like show that $\lambda(A_n \cap A_k) = 0$ implies that $A_n \cap A_k = \emptyset$ but I think it may be not true, for example if I get $A_n = \mathbb{Q}$ and $A_k = \mathbb{R}$

Anyone can help?

Answer 1

HINT: show by induction on $N\ge 0$ that $$\lambda \left(\bigcup_{k=0}^N A_k\right) = \sum_{k=0}^N \lambda(A_k)$$ Then send $N \to + \infty$, recalling that $\left\{ \bigcup_{k=1}^N A_k\right\}_{N\ge0}$ is an increasing chain of Borel sets.