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Let (X, S $\mu$) be a measure space, and $E_1, E_2, ..., E_n \in S$. For each $m$ index $j \in \{1, 2, ..., n\}$ fixed, let $C_m=\{x \in X: x \in E_j $ for exactly $ m$ index $ j \in \{1, 2, ..., n\}\} $. Prove:

  1. $C_m \in S$ (I already did it)
  2. $\sum_{m=1}^n \mu (E_m)= \sum_{m=1}^n m \mu(C_m)$

I am confused about the second part, I am not sure if induction is the right way because I don´t think $\mu (E_m) = m \mu (C_m)$.

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    Consider the case for 2 sets: it would claim that $\mu(E_1) + \mu(E_2) = \mu(E_1 \Delta E_2) + 2\mu(E_1 \cap E_2)$ can you show this? – 2017-02-19
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    Hint: You can write $C_m$ in terms of the family $E_k$ as follows: $$C_n=\bigcup\{E_{k_1}\cap\ldots\cap E_{k_n}\}$$ and then $$C_m=\bigcup\{E_{k_1}\cap\ldots\cap E_{k_m}\}\setminus \bigl(\bigcup_{k=m+1}^nC_k\bigl)$$ – 2017-02-19
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    @HennoBrandsma I can show with 2 sets. But I am having problems with the n+1 in the induction. – 2017-02-19
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    @MariaU What have you tried? It's always best to show work! – 2017-02-20

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