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Let $E=\cup _{i=1}^{\infty} E_i$, where $E_i\subseteq [0,1]$, and $E_1\subseteq E_2\subseteq E_3\subseteq \ldots$.
I want to show that $m^\ast(E_i)\rightarrow m^\ast(E)$, where $m^\ast$ is the Lebesgue outer measure.

My Attempt:
Let $F_1=E_1,F_2=E_2-F_1,~F_3=E_3-(F_1\cup F_2),~F_4=E_4-(F_1\cup F_2 \cup F_3)\ldots.$ Then the $F_i$ are disjoint and $\cup_{i=1}^\infty F_i = \cup_{i=1}^\infty E_i$. But also, $\cup_{i=1}^n F_i = \cup_{i=1}^n E_i$. So we have $$\begin{align*} m^\ast(E) = m^\ast\left(\bigcup_{i=1}^\infty E_i\right) = m^\ast\left(\bigcup_{i=1}^\infty F_i\right) & = \sum_{i=1}^\infty m^\ast (F_i)\\ {\,}{\,} & = \lim_{n\rightarrow \infty}\sum_{i=1}^n m^\ast(F_i)\\ {\,}{\,} & = \lim_{n\rightarrow \infty}m^\ast\left(\bigcup_{i=1}^n F_i\right)\\ {\,}{\,} & = \lim_{n\rightarrow \infty}m^\ast\left(\bigcup_{i=1}^n E_i\right).\\ \end{align*} $$ Now I'm stumped, because I can't get the last equality to be $\lim_{n\rightarrow \infty} m^\ast(E_n)$.
Help!!!

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You've got it, since $\bigcup_{i=1}^{n} E_{i} = E_{n}$.

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  • $\begingroup$ For some reason \bigcup is being rendered as if it were \bigcap. Can anyone explain this? $\endgroup$ – Quinn Culver Feb 1 '12 at 14:33
  • $\begingroup$ It must be something on your side, it renders fine to me. $\endgroup$ – Bruno Stonek Feb 1 '12 at 14:36
  • $\begingroup$ ah! ;) thanks very much...I can't believe, I missed that. $\endgroup$ – Kuku Feb 1 '12 at 14:39

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