TITLE: Continuity of measure
QUESTION [8 upvotes]: 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!!!

REPLY [4 votes]: You've got it, since $\bigcup_{i=1}^{n} E_{i} = E_{n}$.