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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? :)

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    $\begingroup$ It is far from correct. You're assuming properties you still have to prove in order to prove them. It makes very little sense. Look at 5xum's answer below for more details. $\endgroup$ Oct 24, 2018 at 14:24
  • $\begingroup$ Concerning your corrected proof: 1) you should add what $f$ is here. I expect it is the function $\mathbb R\to\mathbb R$ prescribed by $x\mapsto-x$. 2) You write $f(B)$ instead of $f^{-1}(B)$ (as suggested in my answer). Fortunately that works here too because $f$ is a bijection. That however needs to be mentioned because not for every function $f$ we have $f(B)^{\complement}=f(B^{\complement})$. You could save you that trouble by working with preimages instead of images. They are more cooperative. $\endgroup$
    – drhab
    Oct 24, 2018 at 15:34
  • $\begingroup$ Concerning the first and second bullet I would like to see an argumentation for $\varnothing=-\varnothing$ and $\mathbb R=-\mathbb R$. Also there you can use preimages of $f$ as shown in my answer. $\endgroup$
    – drhab
    Oct 24, 2018 at 15:36
  • $\begingroup$ @drhab All right thanks a lot! Your answer and comments helped a lot! I have one last question: How do the measurable functions from $(\mathbb R, E) $ to $(\mathbb R, E)$ look like? $\endgroup$
    – StMan
    Oct 24, 2018 at 19:22
  • $\begingroup$ It can be proved that $g$ is measurable iff $\forall x\in\mathbb R[g(-x)=g(x)\text{ or } g(-x)=-g(x)]$. $\endgroup$
    – drhab
    Oct 25, 2018 at 8:59

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much for your observations. I posted a corrected answer above... is it now ok? $\endgroup$
    – StMan
    Oct 24, 2018 at 15:17
  • $\begingroup$ @StMan Nope. In your corrected proof, you use the term $f(B)$. I don't know what $f$ is. You didn't define it. In mathematics, you can't just use terms without defining them. $\endgroup$
    – 5xum
    Oct 25, 2018 at 7:19
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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 $x\mapsto-x$.

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  • $\begingroup$ Thank you very much for your answer. It helped a lot to develop a corrected one. Please see above. $\endgroup$
    – StMan
    Oct 24, 2018 at 15:16

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