Show that if a set $E$ has positive outer measure Show that if $E$ has finite measure and $\epsilon$ $\gt$ $0$, then $E$ is the disjoint union of a finite number of measurable sets, each of which has measure at most $\epsilon$.
My proof  (My teacher said that part of my proof still wrong, hope someone can help me to solve this proof ): 
Since $u$ is an open subset of $R$, we can be written as a countable disjoint union of open intervals. Suppose first that the number of intervals is infinite, say : $u$ = $\bigcup\limits_{k=1}^\infty$ $I_k$
Where each $I_k$ is an open interval, we have :
$\sum_{k=1}^\infty$ $m^*$($I_k$) = $m^*(u)$ $\lt$ $\infty$
Since this series converges, there is some $n$ such that 
$(*)$ $\sum_{k=1}^\infty$ $m^*$($I_k$) $\lt$ $\epsilon$
We define $A$ by : $\bigcup\limits_{k=1}^n$ $I_k$
If there are only finitely many intervals in $u$, we can label them as $I_1,I_2, . . ,I_n$ and let $I_k=\varnothing$
For $k \ge n+1$, then we define $A=u$ and $(*)$ still hold, now we have $E\A$ ($E$ intersection $A$)$\subseteq A$, so :
$m^*$ ($E\A$) $\le$ $m^*$ ($u\A$)
$=$ $m^*$ ($u$) - $m^*$($A$)
=$\sum_{k=1}^\infty m^*(I_k)- \sum_{k=1}^n m^*(I_k)$
=$\sum_{k=n+1}^\infty m^* (I_k) \lt \epsilon$
On the other hand $A\E$ $\subseteq$ $u\E$, so $m^*(A\E) \le m^*(u\E) \lt \epsilon$
 A: [Probably few appreciate the kind of answer here, where instead of answering the poster's question, one poses a series of related easier and harder questions.  Let me know if you don't like it and I may stop.]
If there is a problem you can't solve then there is a harder problem you can't solve and probably even an easier problem you can't solve.  In this case here are several easier problems which I believe you can solve and which might give you a clue...and a few harder problems to ponder later.

Problem 1.  Let $E=(0,1)$, i.e., the open interval with endpoints $0$ and
  $1$. Show that for any $\epsilon>0$ there is a finite
  pairwise-disjoint collection of measurable sets $\{E_1,E_2,\dots,
  E_n\}$ so that each $E_i$ has Lebesgue measure smaller than $\epsilon$
  and $\bigcup_{i=1}^n E_i=E$.
Problem 2.  Let $E$ be a measurable subset of $(0,1)$, i.e., a subset of the open interval with endpoints $0$ and
  $1$. Show that for any $\epsilon>0$ there is a finite
  pairwise-disjoint collection of measurable sets $\{E_1,E_2,\dots,
  E_n\}$ so that each $E_i$ has Lebesgue measure smaller than $\epsilon$
  and $\bigcup_{i=1}^n E_i=E$.
Problem 3.  Let $E$ be a measurable  set of finite measure. Show that for any $\epsilon>0$ there  are numbers
  $-\infty<a<b<+\infty$ so that the sets $E\cap (-\infty,a)$ and  $E\cap
  (b,\infty )$  are measurable sets each of which has Lebesgue measure
  smaller than $\epsilon$.

After solving the first three problems the OP's problem should seem quite accessible:

Angelo's Problem.  Let $E$ be a measurable subset of finite measure.  Show that, for any $\epsilon>0$, there is a finite pairwise-disjoint
  collection of measurable sets $\{E_1,E_2,\dots, E_n\}$ so that each
  $E_i$ has Lebesgue measure smaller than $\epsilon$   and
  $\bigcup_{i=1}^n E_i=E$.

Now let's have more fun.

Harder Problem.  Let $E$ be a measurable subset of finite measure equal to $1$ and let $n$ be a positive integer.  Show that  there is a finite pairwise-disjoint
  collection of measurable sets $\{E_1,E_2,\dots, E_n\}$ so that each
  $E_i$ has the same  Lebesgue measure    and
  $\bigcup_{i=1}^n E_i=E$.

This problem says that at a birthday party with a one pound, one-dimensional, cake you can split into $n$ equal weight pieces so no-one is shortchanged.
This suggests a much more interesting problem (which I quote from Real Analysis (BBT) Exercise 2:13.8 (Liaponoff's theorem):

Cake Problem: Given a cake with $k$ ingredients (e.g., butter, sugar, chocolate,   garlic, etc.) each nonatomic and of unit mass and
  mixed together in any "reasonable" way, it is possible to cut the cake
  into $k$ pieces such that each of the pieces contains its share of
  each of the ingredients.

A: Notice that for any subset $E$ of reals with finite measure, and $\epsilon>0$ there exists a $K \subset E$ compact such that, $m(E\setminus K)<\epsilon$. The rest is up to you. 
