Construction of sigma-algebras: a counterexample

[psie]

TL;DR

I'm studying the construction of a $\sigma$-algebra starting from a collection of sets $\mathcal E\subset \mathcal P(X)$, e.g. the open sets of $\mathbb R$ to obtain the Borel $\sigma$-algebra. The way to obtain the $\sigma$-algebra this way turns out to be somewhat complicated.

Consider a set $X$ and family of sets $\mathcal E\subset\mathcal P(X)$. Let $\mathcal E_1=\mathcal{E}\cup\{E^c:E\in\mathcal E\}$ and then for $j>1$ define $\mathcal E_j$ to be the collection of all sets that are countable unions of sets in $\mathcal E_{j-1}$ or complements of such. Let $\mathcal E_\omega=\bigcup_1^\infty\mathcal E_j$. One can verify that $\mathcal E_\omega$ is closed under complements, but not under countable unions apparently. The claim is that if $E_j\in\mathcal E_j\setminus\mathcal E_{j-1}$ for each $j$, there's no reason for $\bigcup_1^\infty E_j$ to be in $\mathcal E_\omega$. Why not?

I don't see why $\bigcup_1^\infty E_j\in\mathcal E_\omega$ should or shouldn't hold. Do you know of any concrete counterexample (preferably something that's not too hard to understand), where $\bigcup_1^\infty E_j\in\mathcal E_\omega$ is simply not true? I think that would be very helpful.

 

[fresh_42]

Consider a set $X$ and family of sets $\mathcal E\subset\mathcal P(X)$. Let $\mathcal E_1=\mathcal{E}\cup\{E^c:E\in\mathcal E\}$ and then for $j>1$ define $\mathcal E_j$ to be the collection of all sets that are countable unions of sets in $\mathcal E_{j-1}$ or complements of such. Let $\mathcal E_\omega=\bigcup_1^\infty\mathcal E_j$. One can verify that $\mathcal E_\omega$ is closed under complements, but not under countable unions apparently. The claim is that if $E_j\in\mathcal E_j\setminus\mathcal E_{j-1}$ for each $j$, there's no reason for $\bigcup_1^\infty E_j$ to be in $\mathcal E_\omega$. Why not?

I have an idea but struggle to convert it into an example; mainly because of the nasty complements that are too big. Perhaps you can construct one. The idea is taken from integrals. It is a phenomenon called "vanishing mass at infinity".

Let $I_E$ be the indicator function on $E\subseteq \mathbb{R}$ and $f_n=I_{[n,n+1]}-I_{[n+1,n+2]}.$ Then $\sum_{k=0}^n f_k=I_{[0,1]}-I_{[n+1,n+2]}$ and $\sum_{k=0}^\infty f_k=I_{[0,1]}.$ Hence
$$
\sum_{k=0}^\infty \left(\int_\mathbb{R}f_k(x)\,dx\right)=0\neq 1=\int_\mathbb{R}\left(\sum_{k=0}^\infty f_k(x)\right)\,dx
$$
The sum can be interpreted as a countable union. Therefore we sum up zeros on the left but the equalizing mass on the right has vanished to infinity so we are left with the initial interval of length one.

 

[psie]

The sum can be interpreted as a countable union. Therefore we sum up zeros on the left but the equalizing mass on the right has vanished to infinity so we are left with the initial interval of length one.

Hmm, how do you interpret the sum as a union? And what is your $\mathcal E$ and $E_j$? I don't think it matters much, but I'm changing the intervals to half-open ones so we get disjoint intervals. Then $$\sum_{k=0}^\infty \left(\int_\mathbb{R}f_k(x)\,dx\right)=\sum_{k=0}^\infty \big(\mu([n,n+1))-\mu([n+1,n+2))\big).$$ Now, there's an identity that reads $\mu(B\setminus A)=\mu(B)-\mu(A)$, but this only works if $A\subset B$ and $\mu(A)<\infty$. $A\subset B$ does not hold in this case, so I'm not sure how to interpret the sum as a union.

 

[fresh_42]

Hmm, how do you interpret the sum as a union? And what is your $\mathcal E$ and $E_j$? I don't think it matters much, but I'm changing the intervals to half-open ones so we get disjoint intervals. Then $$\sum_{k=0}^\infty \left(\int_\mathbb{R}f_k(x)\,dx\right)=\sum_{k=0}^\infty \big(\mu([n,n+1))-\mu([n+1,n+2))\big).$$ Now, there's an identity that reads $\mu(B\setminus A)=\mu(B)-\mu(A)$, but this only works if $A\subset B$ and $\mu(A)<\infty$. $A\subset B$ does not hold in this case, so I'm not sure how to interpret the sum as a union.

I admitted that I struggled with it.

I thought about $X=\left\{f:\mathbb{R}\to \mathbb{R}\,|\,\int_\mathbb{R}f(x)\,dx =0\right\}$ or $X=\cup_n\;\{[n,n+1]\,|\,n\in \mathbb{N}_0\}$ or something but the complements got out of control. And I had no good idea of how to transform the intervals $[n,n+1]$ into sets $\mathcal{E}_n.$

I transformed the conditions into formulas as follows, assuming complement always means complement in $X$:
\begin{align*}
\mathcal E_1&=\mathcal{E}\cup\{X/E\, : \,E\in\mathcal E\}\\
\mathcal E_j&=\bigcup_{n\in \mathbb{N}}\{E_n\in \mathcal{E}_{j-1}\ \}\cup \bigcup_{n\in \mathbb{N}}\{X/E_n\in \mathcal{E}_{j-1}\ \} \\
\mathcal{E}_\omega&=\bigcup_{n\in \mathbb{N}}\mathcal{E}_n=
\bigcup_{n\in \mathbb{N}}E_n\cup\bigcup_{n\in \mathbb{N}}(X/E_n)\\
A&=\bigcup_{n\in \mathbb{N}}\{A_n\in \mathcal{E}_n/\mathcal{E}_{n-1}\ \}\stackrel{?}{\not\in}_{i.g.}\mathcal{E}_\omega
\end{align*}
I have chosen $A$ instead of $E$ in order to distinguish it from the many $Es$ you already used in the definition of the $\mathcal{E}_j.$ My thought was that
\begin{align*}
E\in A&\Longrightarrow E\in A_n\text{ for some }n\\
&\Longrightarrow E \in \mathcal{E}_n\setminus\mathcal{E}_{n-1}\\
&\Longrightarrow E \in \mathcal{E}_{m} \vee X\setminus E \in \mathcal{E}_{m} \text{ for some }m<n\\
&\phantom{\Longrightarrow }(E=X\setminus E_m \Rightarrow X\setminus E=E_m\in \mathcal{E}_m)\\
&\text{ AND }\\
&\phantom{\Longrightarrow }E \not\in \mathcal{E}_{m} \wedge X\setminus E \not\in \mathcal{E}_{m} \text{ for any }m<n-1\\
&\Longrightarrow E\in \mathcal{E}_{n-1} \vee X\setminus E\in \mathcal{E}_{n-1}\\
&\Longrightarrow E\in \mathcal{E}_\omega
\end{align*}
Now, if single sets $E$ are in $\mathcal{E}_\omega$ and it is supposed that $A\not\in \mathcal{E}_\omega$ then the union makes the difference. If my formulas are correct, then $E\in A_n$ means that $E\in \mathcal{E}_{n-1}$ or $X\setminus E\in \mathcal{E}_{n-1}.$ So we successively miss the sets in $\mathcal{E}_n$ by one index and never catch up. That's why I thought of elements vanishing to infinity.

 

[psie]

Thanks for clarifying, @fresh_42. I think I understand your reasoning more and more.

As you may have guessed, this is from Folland's real analysis book, and the way he puts it is that he says:

is $\mathcal E_\omega=\mathcal M(\mathcal E)$? In general, no. $\mathcal E_\omega$ is closed under complements, but if $E_j\in\mathcal E_j\setminus\mathcal E_{j-1}$ for each $j$, there is no reason for $\bigcup_1^\infty E_j$ to be in $\mathcal E_\omega$.

Here $\mathcal M(\mathcal E)$ is the $\sigma$-algebra generated by $\mathcal E$. But note the words "in general". So I suppose it is possible for $\bigcup_1^\infty E_j\in\mathcal E_\omega$.

Consider the following argument. Suppose $\bigcup_1^\infty E_j \in \mathcal E_\omega$. Since $\mathcal E_\omega=\bigcup_1^\infty\mathcal E_j$, there should be a $k$ such that $\bigcup_1^\infty E_j \in \mathcal E_{k}$, but $E_{k+1} \in \mathcal E_{k+1} \setminus \mathcal E_{k}$, i.e. we have $E_{k+1} \notin \mathcal E_{k}$. When can we conclude from this that $\bigcup_1^\infty E_j \notin \mathcal E_{k}$? This would establish a contradiction and hence $\bigcup_1^\infty E_j \notin \mathcal E_\omega$.

EDIT: I think one can show under ZFC+CH that $\mathcal E_\omega=\mathcal M(\mathcal E)$ does indeed hold, though I'm not sure about ZFC or in general.

 

[fresh_42]

Yes, in general means that all counterexamples we try to find may turn out to be examples. I'm not completely convinced and would like to see a counterexample, too. We have $\cup_{j=1}^n E_j \in \mathcal{E}_\omega$ but $\cup_{j=1}^\infty E_j {\not\in}_{i.g.} \mathcal{E}_\omega$ so it must be something that slips away to infinity which is why I thought of these indicator functions: finite sums are $0$ and the sum to infinity yields $1.$ It also reminded me of the other discussion:
$$
1\not\in \cup_{j=1}^n \{0.\underbrace{9\ldots 9}_{j\text{ times}}\} \text{ and }1\in \cup_{j=1}^\infty \{0.\underbrace{9\ldots 9}_{j\text{ times}}\}
$$
However, the complements spoiled such examples. I searched for it in Hewitt/Stromberg which has really many exercises but couldn't find anything. What I found was a formula that might be of help (or not):
$$
\mathcal{E}_\omega=\cup_{j=1}^\infty \mathcal{E}_j=\mathcal{E}_1\cup(\mathcal{E}_2\cap (X\setminus \mathcal{E}_1)) \cup\ldots\cup (\mathcal{E}_j\cap (X\setminus \mathcal{E}_{j-1})\ldots\cap (X\setminus \mathcal{E}_{1}))\cup \ldots
$$
There are so many different ways to construct examples of $\sigma$-algebras, that it is almost impossible to find one that suits as a counterexample since the vast majority of them are nice.