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For any $\{E_n\}$ in $M$ (not necessarily disjoint)

Prove $m(\bigcup E_n) \le \sum m(E_n) $

Where $\bigcup$ and $\sum$ are from $n=1$ to $\infty$.

How would you prove this in sigma algebra. Would you need intervals such as $[0,1)$ in $\mathbb{Q}$ or $\mathbb{R}$?

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This is a property of measures called subadditivity: $$ \mu\left(\bigcup_{k=1}^\infty A_k\right)\leq\sum_{k=1}^\infty \mu(A_k) $$ for any $A_k$ in the domain of definition of $\mu$. This is simply implied by the additivity property $$ \mu\left(\bigcup_{k=1}^\infty B_k\right) = \sum_{k=1}^\infty \mu(B_k) $$ whenever $B_k$ are disjoint.

To see this implication, define $C_1 = A_1$, and $C_k = A_k\setminus C_{k-1}$, then clearly $$ \bigcup_{k=1}^\infty A_k = \bigcup_{k=1}^\infty C_k $$ and $C_k$ are disjoint. Moreover, for any $k$: $\mu(C_k)\leq \mu(A_k)$ since $A_k\subseteq C_k$. As a result, $$ \mu\left(\bigcup_{k=1}^\infty A_k\right) = \mu\left(\bigcup_{k=1}^\infty C_k\right) = \sum_{k=1}^\infty \mu(C_k)\leq \sum_{k=1}^\infty \mu(A_k) $$

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