Set is a sigma-algebra 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? :)
 A: None of your points is proven, except for the first, sort of. Even there, you still need to prove that $-\emptyset=\emptyset$. All other points have major flaws in the proofs, and if I were grading them, I would give you very few or zero points.


  
*
  
*$\mathbb R \in E$ because $\mathbb R^c$ is the empty set and therefore belongs to E
  

There is nothing about complements in the definition of $E$. Your argument is not proof that $\mathbb R\in E$. To prove that a set $A$ is in $E$, you need to prove that $A=-A$. You need to do the same with $\mathbb R$.


  
*
  
*if for $B \in E $ it holds that $B=-B$, then $B=(B^c)^c = -(B^c)^c$. Therefore, $B^c \in E$

This is only proof that if $B\in E$, then $(B^c)^c\in E$. You still need to prove that $B^c\in E$. To do that, you need to prove that $-(B^c)=B^c$.


$E_n = -E_n$ 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, $\cup E_n = - \cup E_n$.

You claim that $\cup E_n = - \cup E_n$, but in fact, this is what you need to prove.
A: Your proof is not correct.
E.g. you state that $\mathbb R\in E$ on base of the fact that its complement is an element of $E$. This conclusion can be drawn if it is known already that $E$ is a $\sigma$-algebra. This however is the fact that must be proved (hence cannot be used yet).
And more things are wrong (see the answer of 5xum).

Let $f:\mathbb R\to\mathbb R$ be a function and let $\mathcal E=\{A\in\wp(\mathbb R)\mid A=f^{-1}(A)\}$.
Then it is immediate that $\varnothing,\mathbb R\in\mathcal E$.
Further if $A\in\mathcal E$ then: $$A^{\complement}=f^{-1}(A)^{\complement}=f^{-1}(A^{\complement})\tag1$$ showing that $\mathcal E$ is closed under complementation.
If $A_{\lambda}\in\mathcal E$ for every $\lambda\in\Lambda$ then $$\bigcup_{\lambda\in\Lambda}A_{\lambda}=\bigcup_{\lambda\in\Lambda}f^{-1}(A_{\lambda})=f^{-1}\left(\bigcup_{\lambda\in\Lambda}A_{\lambda}\right)\tag2$$
showing that $\mathcal E$ is closed under arbitrary unions.
In $(1)$ and $(2)$ it is not difficult to verify the second equality.
This together proves that $\mathcal E$ is a $\sigma$-algebra (even stronger, it is closed under arbitrary unions).
Now apply this for the function $f:\mathbb R\to\mathbb R$ prescribed by $x\mapsto-x$.
