How to think of this problem: $\mathcal{A}_{\mu}$ not necessarily equal to $\mathcal{A}_{\nu}$ when $\mu$ and $\nu$ are finite measures on a measurable space $\left(X,\mathcal{A}\right)$? NOTE: I would like you to tell me if my approach below is correct and answer my qestions.
$\textbf{My thoughts:}$ First, the book suggests me finding an example. I think, I should think of some $\sigma-$algebra $\mathcal{A}$ (say, $\mathcal{B}\left([0,1]\right)$), measures $\mu, \nu$ on it and such a set $A$ that is in, say, $\mathcal{A}_{\mu}$, but not in $\mathcal{A}_{\nu}$. Then $\mathcal{A}_{\mu}$ and $\mathcal{A}_{\nu}$ would not be eaqual. Now, if I write down the definitions
$A \in \mathcal{A}_{\mu}$ means that there exist $E,F \in \mathcal{A}$ such that $E \subseteq A \subseteq F$ and $\mu\left(F-E\right)=0$
$A \in \mathcal{A}_{\nu}$ means that there exist $G,H \in \mathcal{A}$ such that $G \subseteq A \subseteq H$ and $\nu\left(H-G\right)=0$
I see that, I have to find a set $A$ such that, for example, under measure $\mu$ the set $F-E$ is $0$ while under measure $\nu$ there are no such sets $G,H \in \mathcal{A}$ that $\nu\left(H-G\right)=0$ despite $G \subseteq A \subseteq H$. Now, what if I took $E=F=G=H=A \in \mathcal{A}$? Since $\mu$ and $\nu$ are measures, $\mu\left(F-E\right)=0$ and $\nu\left(H-G\right)=0$. Hence, for every set $A \subseteq X$ there exist $E,F,G,H \in \mathcal{A}$ with properties above unless $A$ itself is not measurable under $\mu$ or $\nu$.
$\textbf{My questions:}$ if I took measurable space $\left([0,1],\mathcal{B}\left([0,1]\right)\right)$ and Lebesgue measure $\lambda$, is it true that all sets $A$ in $\mathcal{B}\left([0,1]\right)$ and all subsets of $A$ are Lebesgue measurable? I think so, because the Lebesgue measure is an extension of the Borel measure. If this is true, then let $A=K - $Cantor set. Now what finite measure $\nu$ to choose such that the Cantor set is not measurable under $\nu$? Tell me if what I wrote above was a good thinking and if it was not or if there is a better approach, tell yours.
Hint: Let $X=[0,1]$, $\mu$ be a measure such that not every subset of $X$ is in $\mathcal{A}_\mu$ and $\nu$ is a measure that assigns measure zero to $(0,1]$.