Let's start with basic definitions.

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 $\color{blue}{\text{blue}}?$


2 Answers 2


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 $\Gamma(\mathcal{A})$.

  • $\begingroup$ 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. $\endgroup$ Oct 10, 2015 at 16:35
  • $\begingroup$ I see now. I wrote stronger, but I ment weaker condition. My bad. I am going to change it. $\endgroup$ Oct 10, 2015 at 16:38
  • $\begingroup$ By $\gamma(\mathcal{A})$ I meant your $\Gamma$ construction applied to $\mathcal{A}$; I've edited the answer to be clearer. $\endgroup$ Oct 10, 2015 at 20:12
  • $\begingroup$ 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? $\endgroup$ Oct 10, 2015 at 22:27
  • 1
    $\begingroup$ @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 $\mathcal{A}$, $\Gamma(\mathcal{A})=\gamma(\mathcal{A})$. $\endgroup$
    – Ramiro
    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.


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

  • 1
    $\begingroup$ With this definition, it is not always true that $\Gamma$ is the smallest $\gamma$-algebra containing $\mathcal{A}$; see my answer. $\endgroup$ Oct 10, 2015 at 20:13

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