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Exercise

Let $(X,\Sigma,\mu)$ be a measure space and let $(A_k)_{k \in \mathbb N}$ be a sequence of measurable sets. For each $m \in \mathbb N$, we define $B_m$ as the subset of all points in $X$ which belong to at least $m+1$ sets of the sequence. Prove that $B_m$ is measurable for each $m$ and that $$m\mu(B_m) \leq \sum_{k \geq 1} \mu(A_k)$$

First of all, for $m$ fixed, I am trying to write $B_m$ as countable union/intersection (or combination) of measurable sets, but I am having some trouble doing this. What I thought was writing $$B_m=\bigcup_{j=m+1}^{\infty} \bigcap_{k=1}^{j} A_k$$ but I know this is not quite what I want since this is "all the points that belong to each of the first $j$ sets for all $j \geq m+1$, I don't know how to fix this in order to get $B_m$ properly expressed. As for the inequality I suppose once I can express $B_m$ in terms of the sets of the sequence I will know how to show it.

I would appreciate suggestions, thanks in advance.

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Let $$ C := \{F \subset \mathbb{N} : |F| = m+1\} $$ Then $C$ is countable, and $$ B_m = \bigcup_{F \in C} \bigcap_{i\in F} A_i $$ Thus, $B_m$ is measurable. Perhaps with this you can try proving the inequality?

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  • $\begingroup$ No, your $C$ should be defined as $\{F \subset \mathbb{N}: |F| \geq m+1\}$, and this is no longer a countable set since $\mathbb{N}$ contains uncountably many subsets of infinite cardinality. $\endgroup$
    – shalop
    Commented May 14, 2015 at 3:16
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    $\begingroup$ A priori one does need all those sets, but a simple containment check will show you that you only need those with cardinality exactly $m+1$ (Note that any subset of cardinality $\geq m+1$ contains a set of cardinality $m+1$ and so the intersection over the larger set is contained in the intersection over the smaller set) $\endgroup$
    – Prahlad Vaidyanathan
    Commented May 14, 2015 at 3:25
  • $\begingroup$ Oh okay, I see... $\endgroup$
    – shalop
    Commented May 14, 2015 at 3:30

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