# Modifying definition of $\sigma$-algebra

Def. Given a set $X$ we say that $\Sigma\subset 2^X$ is a $\sigma$-algebra, if $X\in\Sigma,$ $\Sigma$ is closed under complementation and countable unions.

Def. Given a family $\mathcal A$ of subsets of $X,$ we say that $\sigma$-algebra $\sigma(\mathcal A)$ is the $\sigma$-algebra generated by $\mathcal A$, if it is the smallest $\sigma$-algebra containing $\mathcal A.$

Asaf Karagila said in this answer, that $\sigma(\mathcal A)$ has descriptive, but transfinite form. It is

• $\Sigma^0_0=\Pi^0_0=$ finite intersections from $\mathcal{A}$
• For countable ordinals $\alpha$ let:

$$\Sigma^0_\alpha=\{\bigcup_{i\in\mathbb N} A_i\mid A_i\in\bigcup_{\beta<\alpha}\Pi^0_\beta\},\quad \Pi^0_\alpha = \{X\setminus A\mid A\in\Sigma^0_\alpha\},\quad \Delta^0_\alpha=\Sigma^0_\alpha\cap\Pi^0_\alpha$$ and then we define $\Delta=\bigcup_{\alpha<\omega_1} \Delta^0_\alpha.$

I heard of slightly different approach (I colored differences in $\color{green}{\text{green}}$), i.e.

• $\color{green}{\Delta_0=\mathcal{A}\cup\{\emptyset\}}$
• For countable ordinals $\alpha$ let:

$$\Sigma_\alpha=\{\bigcup_{i\in\mathbb N} A_i\mid A_i\in\bigcup_{\beta<\alpha}\color{green}{\Delta}_\beta\},\quad \Pi_\alpha = \{X\setminus A\mid A\in\Sigma_\alpha\},\quad \Delta_\alpha=\Sigma_\alpha\color{green}{\cup}\Pi_\alpha$$

and then we define $\Delta=\bigcup_{\alpha<\omega_1} \Delta_\alpha.$

In both cases the claim is that $\Delta=\sigma(\mathcal A).$

I have no clue how someone is able to proof that, particularly the inclusion $\Delta\subset\sigma(\mathcal A),$ but I believe it can be done and I am willing to assume that $\Delta=\sigma(\mathcal A)$ in both cases above.

From now on, if I say transfinte induction, I mean the second one, i.e. the one with $\color{green}{\text{green}}$ elements.

I would like to create a new object $\Gamma,$ which I get from transfinite induction, but restricted to the finite ordinals, i.e. I just simply do normal induction and define $\Gamma=\bigcup_{n<\infty}\Delta_n.$ To be even more explicite, let's actually define it using good old normal induction. So we set

• $\Delta_0=\mathcal{A}\cup\{\emptyset\}$
• For every $n\in\mathbb{N}:$

$$\Sigma_n=\{\bigcup_{i\in\mathbb N} A_i\mid A_i\in\bigcup_{k<n}\Delta_k\},\quad \Pi_n = \{X\setminus A\mid A\in\Sigma_n\},\quad \Delta_n=\Sigma_n\cup\Pi_n$$ and then we define $\Gamma=\bigcup_{n<\infty}\Delta_n.$

For me $\Gamma$ is big enought. But since $\Delta=\sigma(\mathcal A)$ and $\Gamma$ looks smaller than $\Delta$ it is likely that $\Gamma\neq\sigma(\mathcal A).$

Unfortunetely $\Gamma$ seems not to be even a $\sigma$-algebra, for the same reason that $\Delta$ is in fact $\sigma$-algebra. (See Arturo Magidin's answer) It fails to be closed under countable unions. But it fails, due to considering diagonal-like elements (and actually this is the reason why I define $\Gamma$ in such way).

What is the idea! (don't be scared in first reading)

I would like to define two new objects called $\gamma$-algebra and $\gamma$-algebra generated by $\mathcal A.$ $\gamma$-algebra is defined similar to $\sigma$-algebra, but with $\color{blue}{\text{different (weaker) conditions}}$. $\gamma$-algebra generated by $\mathcal A$ is smallest such $\gamma$-algebra and it has descriptive form, which equals $\Gamma.$

In other words, I would like to end up with the following situation:

Def. Given a set $X$ we say that $\Omega\subset 2^X$ is a $\gamma$-algebra, if $X\in\Omega,$ $\Omega$ is closed under complementation and $\color{blue}{\text{some conditions I ask you to invent}}$.

Def. Given a family $\mathcal A$ of subsets of $X,$ we say that $\gamma$-algebra $\gamma(\mathcal A)$ is the $\gamma$-algebra generated by $\mathcal A$, if it is the smallest $\gamma$-algebra containing $\mathcal A.$

Thm. Let $\mathcal A$ be a family of subsets of $X.$ If we define $\Gamma$ as above, then $$\Gamma=\gamma(\mathcal A).$$

The question

What conditions should be in $\color{blue}{\text{blue}}?$

No such blue conditions exist. Write $\Gamma(\mathcal{A})$ for your construction $\Gamma$ applied to a starting set $\mathcal{A}$. If there was such a notion of $\gamma$-algebra, then the operation $\gamma$ would satisfy $\gamma(\gamma(\mathcal{A}))=\gamma(\mathcal{A})$ (since $\gamma(\mathcal{A})$ is a $\gamma$-algebra, and hence is the smallest $\gamma$-algebra containing $\gamma(\mathcal{A})$). Also your theorem would say that $\gamma(\mathcal{A})=\Gamma(\mathcal{A})$ for all $\mathcal{A}$, and so we must have $\Gamma(\Gamma(\mathcal{A}))=\Gamma(\mathcal{A})$ for all $\mathcal{A}$. But this is not true, since $\Gamma(\mathcal{A})$ is typically not closed under countable unions, whereas $\Gamma(\Gamma(\mathcal{A}))$ contains all countable unions of elements of $\Gamma(\mathcal{A})$.

• I do not see why $\gamma(\gamma(A))$ contains all countable unions of elements of $\gamma(A).$ If we demand the condition that $\gamma$-algebra is closed under countable unions, then that would be the case. But the point of my question is to invent some alternative to closeness of countable unions. Oct 10, 2015 at 16:35
• I see now. I wrote stronger, but I ment weaker condition. My bad. I am going to change it. Oct 10, 2015 at 16:38
• By $\gamma(\mathcal{A})$ I meant your $\Gamma$ construction applied to $\mathcal{A}$; I've edited the answer to be clearer. Oct 10, 2015 at 20:12
• You are just right. But how could you be so sure with the statement: "No such blue conditions exist"? Is there any reason that none can come up with some bright idea and state contidion in blue in simple way? Oct 10, 2015 at 22:27
• @FallenApart The Eric's argument is correct. Let me put in other words. If there is a concept of $\gamma$-algebra and we define $\gamma(\mathcal{A})$ to be the smallest $\gamma$-algebra containing $\mathcal{A}$, then, as direct consequence of the definition, we have $\gamma(\gamma(\mathcal{A}))= \gamma(\mathcal{A})$. However, $\Gamma(\Gamma(\mathcal{A}))\neq \Gamma(\mathcal{A})$. So, it is not possible that FOR ALL $\mathcal{A}$, $\Gamma(\mathcal{A})=\gamma(\mathcal{A})$. Oct 10, 2015 at 22:49

I have something. So the condition is following:

$\Omega$ admits a countable filtration $\Omega=\bigcup_{n<\infty}\Delta_n$ closed under countable bounded (with respect to filtration) unions, i.e. for every $n\in\mathbb{N}$ and every $A_1,A_2,\dots\in\Delta_n$ (these are bounded elements) $$\bigcup_{i=1}^\infty A_i\in\Omega$$ Condition $\bigcup_{i=1}^\infty A_i\in\Delta_{n+1}$ would work as well, but the upper one is weaker.

The key idea is to forbid taking diagonal-like elements in countable unions. Boudedness do the job.

Edit

Eric is correct. In such definition, $\Gamma$ wouldn't be the smallest $\gamma$-algebra containing $\mathcal A.$

• With this definition, it is not always true that $\Gamma$ is the smallest $\gamma$-algebra containing $\mathcal{A}$; see my answer. Oct 10, 2015 at 20:13