Are supersets of non-empty measurable sets measurable? Challenging conventional wisdom question
Let $(S, \Sigma, \mu) = (\mathbb{R}, \mathscr{M}(\mathbb{R}), \lambda)$, where $\mathscr{M}(\mathbb{R})$ is the collection of all $\lambda$-measurable subsets of $\mathbb{R}$
If $A \in \Sigma, A \ne \emptyset$ and $B \subseteq A$, it does not follow that $B \in \Sigma$.
However, if $C \supseteq A$, do we have $C \in \Sigma$?
I guess we can have $C = A \cup V$ where $V \notin \Sigma$ (eg Vitali set) and $V \subsetneq A$, right?
 A: Take your favorite measurable subset $A$ of $(0, 1)$, and favorite non-measurable subset $B$ of $(2, 3)$. Then $A\cup B$ is non-measurable, but is a superset of $A$.

Note that the example you give in your question, while in the right direction, doesn't quite work - if we take $V$ to be a Vitali subset of $(0, 1)$, and $A=(0, 1)\setminus \{x\}$ for some $x\in V$, then $A\cup V$ is measurable. It's not enough just to assume that $V\not\subseteq A$, you have to assume that $V$ is "sufficiently separate from" $A$ in some sense. One very brute force way of accomplishing this is to have $A$ and $V$ live in disjoint intervals, which is the approach I've taken above.
A: If you know that the $\sigma$-algebra of Lebesgue measurable sets is a proper subset of $\mathcal{P}(\Bbb R)$, then this is easy to prove.
The $\sigma$-algebra of Lebesgue measurable sets is complete, which means all subsets of sets of measure $0$ are measurable.
Let $A$ be any non-empty subset of measure $0$, e.g., $A = \{1,2,3 \}$.  Then if every superset of $A$ were also measurable, then every subset of $\Bbb R$ is measurable, which would mean the $\sigma$-algebra of Lebesgue measurable sets is equal to $\mathcal{P}(\Bbb R)$, a contradiction.
