I have a question about Lebesgue measure Suppose $A$ is Lebesgue measurable, show that for each $\epsilon>0$, there exists an open set, $O$, such that $A\subset O$, and $\lambda(O-A)<\epsilon$. (note: $\lambda$ is the Lebesgue measure in $\mathbb R$).
If we just deal with the finite case first, then $\lambda(A)<\infty$, since $A\subset\mathcal{M}$, we have that $\lambda(A)=\lambda^{*}(A) = \inf\{\sum_{n=1}^\infty l(I_n) \ | \ \{I_n\}_{n=1}^\infty\text{ is a sequnce of open intervals s.t. }A\subset\cup_{n=1}^\infty I_n\}$, So I guess my question is how do I reference this sequence of open intervals? Can I just say that we have the "smallest" such $I_n$(call it $O$) that covers A, and the sum of the lengths of the intervals is the outer measure of A so, $\lambda(O-A)=\lambda(O)-\lambda(A)=\lambda^*(O)-\lambda^*(A)=0<\epsilon$. Any comments would greatly be appreciated.
 A: Assume $\lambda(A)<\infty$.
First off, we know that the Lebesgue measure is only a restriction of the outer measure to measurable sets, i.e. as you also wrote $\lambda(A)=\lambda^*(A)$ for measurable $A$. We do not need to prove that. This is the definition of $\lambda$.
You cannot pick a smallest covering by intervals because the infimum may not be taken. 
This is like if you take the infimum of the set $\{1/n\,:\,n\in\mathbb{N}\}$ (which is $0$): $1/n$ will get really, really close to $0$ as $n$ gets really, really large, but it will never quite reach zero. The minute you pick one of the $1/n$'s and declare it the "smallest" you lose because you're still infinitely many $n$'s far from $0$. But you can pick an $\epsilon>0$ and decide that's how far you want to approximate $0$ and then the definition of the infimum gives you an $n$ such that $1/n-0<\epsilon$.
Now let us do the same here. Choose $\epsilon>0$. By the definition of the infimum, there exists a sequence of open intervals $(I_n)_n$ covering $A$ such that 
$$\sum_{n=1}^\infty l(I_n)-\lambda(A)<\epsilon$$
Now write $O=\bigcup_{n=1}^\infty I_n$. $O$ is open as union of open intervals. By the subadditivity of the Lebesgue measure
$$\lambda(O)\le \sum_{n=1}^\infty l(I_n)$$
All of this we could have done also without assuming measurability of $A$, since also the outer measure $\lambda^*$ is subadditive. 
The crucial step now, which you also already did, uses measurability of $A$ (this follows literally from the classical Carathéodory definition of measurability, though I don't know if you used that to define measurability, so look it up if not!):
$$\lambda(O)=\lambda(A)+\lambda(O- A)$$
This implies 
$$\lambda(O-A)=\lambda(O)-\lambda(A)\le \sum_{n=1}^\infty l(I_n) - \lambda(A)<\epsilon$$
which is what we wanted to prove. The interesting part is that you can actually also use this property (called outer regularity of the Lebesgue measure) to define measurability and derive the Carathéodory definition back from it (there may be some subleties with all sets of measure $0$ being measurable or not, but let's not go into this now).
Edit: 
All this was for the case $\lambda(A)<\infty$. If $\lambda(A)=\infty$, we need to choose an exhaustion of $A$ by compact sets, i.e.
$$A_k=\{x\in A\,:\,|x|\le k\}$$
then $\lambda(A_k)<\infty$ and $\bigcup_{k=1}^\infty A_k=A$. We apply the above argument to obtain a sequence of open sets $(O_k)_k$ with the property
$$\lambda(O_k-A_k)\le \frac{\epsilon}{2^k}$$
Now define $O=\bigcup_{k=1}^\infty O_k$. Then
$$\lambda(O-A)=\lambda\left(\bigcup_{k=1}^\infty (O_k-A)\right)\le \sum_{k=1}^\infty \lambda(O_k-A)\le \sum_{k=1}^\infty \lambda(O_k-A_k)\le \epsilon$$
