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 $m^*(A\E) \le m^*(u\E) \lt \epsilon$
Best Answer
[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.
After solving the first three problems the OP's problem should seem quite accessible:
Now let's have more fun.
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):