$A \subset \mathbb{R} $ is measurable, prove that $-A=\{x : -x \in A\}$ is measurable. $A \subset \mathbb{R} $ is measurable, prove that $-A=\{x : -x \in A\}$ is measurable.
It is more than obvious that $-A$ is measurable, but I am sure that I am not supposed to say :"$-A$ is just $A$ displaced on the real line". My Question is : if I show that for each $\epsilon > 0$ there exists an open set $O$ containing $-A$ such that $m(O \setminus -A) < \epsilon$, does this imply that $-A$ is measurable ?
 A: The set $A\subset\mathbb R$ is measurable if and only if $\chi_A$ is a measurable function. But is $f(x)$ is measurable, so is $h(x)=f(-x)$, as $h=f\circ\varphi$, where $\varphi$ is a continuous function. Note that $\chi_{A}(-x)=\chi_{-A}(x)$, and thus $-A$ is also a measurable set.
A: I reckon you are thinking just of the sigma-algebra of the Borelsets of
$\mathbb{R}$. For that look at the other answers. More general: 
$\mathcal{A}:=\left\{ \emptyset,\left(-\infty,0\right],\left(0,\infty\right),\mathbb{R}\right\} $
is also a sigma-algebra on $\mathbb{R}$. 
Here $A:=\left(-\infty,0\right]$
is measurable (i.e. is an element of $\mathcal{A}$) but $-A=\left[0,\infty\right)$
is not.
A: As usual, consider the family of measurable sets $A$ such that $-A$ is measurable. This is a sigma-algebra which contains every interval hence it contains the sigma-algebra generated by the family of intervals, which is...
A: Let $A\subseteq \Bbb R$ be measurable. By your definition (whatever $m^*$ means), for any $B\subseteq \Bbb R$
$$m^*(-B)=m^*(-B\cap A)+m^*(-B\cap A^c).$$
I'm betting that $m^*(-X)$ will always equal $m^*(X)$ in some trivial way. But then
\begin{align*}
m^*(B)=m^*(-B)&=m^*(-B\cap A)+m^*(-B\cap A^c) \\
&=m^*(B\cap -A)+m^*(B\cap (-A)^c)
\end{align*}
