# 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? :)

• 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. Oct 24, 2018 at 14:24
• 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. Oct 24, 2018 at 15:34
• 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. Oct 24, 2018 at 15:36
• @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? Oct 24, 2018 at 19:22
• It can be proved that $g$ is measurable iff $\forall x\in\mathbb R[g(-x)=g(x)\text{ or } g(-x)=-g(x)]$. Oct 25, 2018 at 8:59

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.

• Thank you very much for your observations. I posted a corrected answer above... is it now ok? Oct 24, 2018 at 15:17
• @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.
– 5xum
Oct 25, 2018 at 7:19

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 $$x\mapsto-x$$.

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