Let $A \subset \mathbb R$ be denoted as: $-A := \left\{-x : x \in A\right\}$.

Why is $E:= \{ A \subset \mathbb R : A = -A\}$ a $\sigma$-algebra on $\mathbb R$? 

Here is my proof:

* the empty set belongs to $E$ because it holds true that $- \emptyset = \emptyset $
* $\mathbb R \in E$ because $\mathbb R^c$ is the empty set and therefore belongs to $E$
* if for $B \in E $ it holds that $B=-B$, then $B=(B^c)^c = -(B^c)^c$. Therefore, $B^c \in E$
* if $E\_n = -E\_n$ for all $n$, then $\cup E\_n = - \cup E\_n$. If $E\_n^c = - E\_n^c$ for some m, then, since $E\_m \subset \cup E\_n$, we have $(\cup E\_n)^c \subset A\_m^c$ and $-(\cup E\_n)^c \subset E\_m^c$. Then $(\cup E\_n)^c = -(\cup E\_n)^c$ and, by my third point, $\cup E\_n = - \cup E\_n$.

Is this proof correct? Or what would you write in other terms / words? And how?

EDIT:

corrected proof:

* the empty set belongs to E since $\emptyset$ as well as $- \emptyset$ are in E by definition [is this ok?]
* $\mathbb R$ is in E since $\mathbb R$ as well as $- \mathbb R$ are in E by def.
* if $B \in E$ then: $B^c = f(B)^c = f(B^c) = -B^c$ --> E is closed under complements
* if $B\_a \in E $ for all $a \in \mathbb N$, then: $\cup B\_a = \cup f(B\_a) = f(\cup B\_a) = - \cup B\_a$ showing that E is closed under arbitrary unions.

Is this acceptable? :)