Let $(X,M,\mu)$ be a measurable space with $0<\mu(X)<\infty$ Let $(A_n)$be a sequence of measurable sets and $k>0$ a natural number, and suppose that every $x \in X$ is conteined in at most $k$ sets of the sequence.
Prove:
(i) $\sum_n \mu(A_n) \leq k\mu(X)$ (ii) if every $x$ is contained in infinite elements of the sequence, then $\sum_n \mu(A_n)= \infty$
What i tried by far: That seems easy to see, but cant find a good way to prove. $B_j$:= {x | x is contained j times in (A_n) } and my goal was proving $\sum_{j=1}^k \mu(B_j )j \ge \sum_n \mu(A_n)$ But I dont think those $B_j$ are measurable
Let $1_{A_n}$ denotes the indicator function of $A_n$. Then we have $$ \sum_n \mu (A_n) = \sum_n \int 1_{A_n} = \int \underbrace{\sum_n 1_{A_n}}_{\le k} \le k \mu(X). $$
You are allowed to switch the (infinite) sum with the integral, as each $1_{A_n}$ is non-negative. If you want to be pedantic, go for the partial sums.
You can adapt this idea also for the second part of the question. Just estimate the interval from below.