Measurable sets on [0,1] Hi guys I know this is should be easy, but I need to brush up my analysis
I have two measurable sets say $E,K$on $[0,1]$ and that $m(E)=1$ and we want to show is that $m(E \cap K)= m(K)$
My idea was to argue that $m(E \cup K)= m(E) +m(K)- m(E \cap K)$ thus we get that $1=1+m(K)-m(E \cap K)$ and we are done. Does this seem true?
 A: It is true that $m(E \cup K) = m(E) + m(K) - m(E \cap K)$ but you have to prove it.  
I would instead note the following facts:


*

*$K = (K \cap E) \cup (K \setminus E)$ as a disjoint union.

*$K \setminus E \subset [0,1] \setminus E$.  

*$[0,1] = E \cup ([0,1] \setminus E)$ as a disjoint union.
A: For visitors in the future, here is a more thorough explanation based off of Nate Eldredge's answer.
Here, I presuppose that if $K,E\subseteq[0,1]$ are measurable, then $K\cap E$ (and less importantly $K\cup E$) are measurable. Now, we can express $K$ as $$K=\underbrace{(K\cap E)}_\text{measurable}\cup\underbrace{(K\setminus E)}_\text{measurable?}.$$ It is obvious by Venn diagram that $(K\cap E)$ and $(K\setminus E)$ are disjoint.
Next, notice that since $[0,1]$ is an interval and thus measurable and that $E$ is measurable by assumption. Then since $E\subseteq[0,1]$, we may say that $$m([0,1]\setminus E) = m[0,1] - m(E) = 1-1 = 0.$$
Now, observe that $K\setminus E\subseteq\underbrace{[0,1]\setminus E}_\text{measurable}$. It is obvious that since $[0,1]\setminus E$ has measure zero, that any subset is measurable and also has measure zero. Thus $K\setminus E$ is measurable and $m(K\setminus E) = 0$.
Therefore, since both $(K\cap E)$ and $(K\setminus E)$ are both disjoint and measurable, we can invoke the following (since $K$ is already assumed to be measurable): $$\begin{align}m(K) &= m[(K\cap E)\cup(K\setminus E)] \\ &= m(K\cap E) + m(K\setminus E) \\ &= m(K\cap E) + 0 \\ &= m(K\cap E).\end{align}$$
