Let $(X, \Sigma, \mu)$ be a measure space and $\{A_k\}_{k \geq 1}$ be a sequence of measurable sets. Consider now
$$ B_q = \{x \in X : \exists i_1 < \dots < i_q \text{ with } x \in A_{i_j}\} \quad (\forall q \in \mathbb{N}) $$
that is, the set of elements of $x$ that lie on at least $q$ sets of the sequence. This sets are measurable since $\phi = \sum_k\chi_{A_k}$ is a measurable function and $\mu(B_q) = \mu(\{\phi^{-1} \geq q\})$. Now, I'm trying to prove the following bound
$$ q\cdot\mu(B_q) \leq \sum_{k\in\mathbb{N}}\mu(A_k) $$
I've seen already that the sets $B_q$ are decreasing, and intuitively I presume I should show that some expression possibly involving $B_1, \dots B_q$ or $B_q \dots B_{2q}$ (for example by writing $q\mu(B_q) = \sum_{i=1}^q\mu(B_q)$ and then bounding the latter) is contained in $\bigcup_{n \in \mathbb{N}}A_n$. So far I've had no luck when attempting to do so. Any hints?
As you said yourself, $B_q \subset \{x \,:\, \phi (x) \geq q\}$. Hence, $q \cdot 1_{B_q} \leq \phi$, and thus $$q \cdot \mu(B_q) =\int q \cdot 1_{B_q} d\mu \leq \int \phi d \mu =\sum_k \mu(A_k).$$