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Suppose $X, M, \mu$ is a measure space and $A_1, A_2, A_3, ...$ are sets in $ M$ such that each point of $X$ belongs to no more than d of the $A_k$ 's. then $\sum^{\infty}_{k=1} \mu(A_k) \le d \mu (\bigcup^{\infty}_{k=1} A_k)$
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If $A_k$ are pairwise disjoint, i.e. $A_k \cap A_i = \varnothing$ if $i \neq k$, then you have $$ \sum_k \mu (A_k) = \mu (\bigcup_k A_k)$$ Next consider the case where each point $x$ in $X$ is in exactly $d$ of the $A_k$. In this case you count the measure of each $A_k$ $d$-times too many in the sum, hence $$ \sum_k \mu (A_k) = d \mu (\bigcup_k A_k)$$ Now you see that if each point is in at most $d$ of the $A_k$, you count the measure of each $A_k$ at most $d$-times too many, hence $$ \sum_k \mu (A_k) \leq d \mu (\bigcup_k A_k) $$ |
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