Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!!!

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

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

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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