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 $\lim_{n\rightarrow \infty} m^\ast(E_n)$.


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.