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Let us call a subset $E$ of $[0,1]$ Lebesgue measurable if $\lambda^*(E) + \lambda^*([0,1]\setminus E) = 1$, where $\lambda^*$ is the outer measure. How can we derive from this definition the fact that countable unions of measurable sets are measurable? (I have already proved this fact for disjoint unions)

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  • $\begingroup$ Prove $E\F$ is measurable where $E,F$ are measurable. $\endgroup$ – Rubertos Apr 10 '15 at 20:46
  • $\begingroup$ @StoneLight If you've already proved the fact for disjoint unions, why not apply that to this general case? If $\{E_n\}$ is a sequence of measurable subsets of $[0,1]$, set $F_1 = E_1$ and $F_n = E_n \cap E_{n-1}^c \cap \cdots \cap E_1^c$ for all $n > 1$. Then the $F_n$ are disjoint measurable subsets of $[0,1]$ such that $\bigcup_{n = 1}^\infty F_n = \bigcup_{n = 1}^\infty E_n$. $\endgroup$ – kobe Apr 10 '15 at 20:49
  • $\begingroup$ Then, prove that $E\cup F$ is measurable (Hint: $E\cup F = (E\setminus F) \cup F$). Then for finite unione. Then prove for a countable sequence of measurable sets $\{E_n\}$. (Hint: define $E'_{n+1}=E_{n+1}\setminus \bigcup_{i=1}^n E_i$.) $\endgroup$ – Rubertos Apr 10 '15 at 20:50

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