In the second paragraph of the proof it says:

"To prove that $\mathcal{B}(\mathbb{R})$ is also generated by the other classes of intervals, it suffices to prove that any interval $]a,b[$ is contained in the sigma algebra corresponding to each class."

I guess there is one more thing one should show: that those sigma algebras do not contain anything that's not in Borel sigma algebra. Is it enough to say that those intervals are contained in Borel sigma algebra? This fact is not trivial and it should be proven I believe.



Let $A$ be the collection of all open intervals. By theorem 9. we know that $\sigma(A) = \mathcal{B}(\mathbb{R})$. Now let $M$ be the collection of intervals of the form $]-\infty, c[$, where $c \in \mathbb{R}$.

The quoted sentence says that in order to show $\sigma(M) = \mathcal{B}(\mathbb{R})$, we need to prove that $A \subseteq \sigma(M)$, thus $\sigma(A) \subseteq \sigma(M)$, thus $\mathcal{B}(\mathbb{R}) \subseteq \sigma(M)$ (from the fact that $\sigma(A) = \mathcal{B}(\mathbb{R})$).

We want to prove that $\sigma(M) = \mathcal{B}(\mathbb{R})$. It can be done by showing that $\mathcal{B}(\mathbb{R}) \subseteq \sigma(M) \subseteq \mathcal{B}(\mathbb{R})$.

So far we have only showed that $\mathcal{B}(\mathbb{R}) \subseteq \sigma(M)$. The book says this is enough. It's not enough - we have to prove $\sigma(M) \subseteq \mathcal{B}(\mathbb{R})$ as well.


Maybe this might help to understand the proof. If $X$ is a set and $\mathcal{A},\mathcal{B} \subset \mathcal{P}(X)$, then if we denote the $\sigma$-algebra generated by $\mathcal{A}$ and $\mathcal{B}$ by $\sigma\mathcal{A}$ and $\sigma\mathcal{B}$ respectively, then we can make the following observations:

If $\mathcal{A} \subset \mathcal{B}$ ,then $\sigma\mathcal{A} \subset \sigma\mathcal{B}$ and

$\sigma\mathcal{A} = \sigma\mathcal{B}$ if and only if $\mathcal{A} \subset \sigma\mathcal{B}$ and $\mathcal{B} \subset \sigma\mathcal{A}$.

Using this, the proof follows easily.

  • $\begingroup$ I understand the proof of theorem 9. My question is related to the second sentence in the picture ("We show that it is also generated by any of the following classes of intervals..."). $\endgroup$ – user4205580 Jan 26 '15 at 11:09
  • $\begingroup$ Since, the $\sigma$-algebra generated by any system of subsets of a set $X$ is the 'smallest' $\sigma$-algebra that contains the system the result follows. $\endgroup$ – sqtrat Jan 26 '15 at 11:11
  • $\begingroup$ Check my update. $\endgroup$ – user4205580 Jan 26 '15 at 11:48
  • $\begingroup$ Now I see what you mean, what you're saying is true, but if you think about it a bit, the other direction is trivial, since $M$ is clearly a subset of $\sigma\mathcal{A}$. $\endgroup$ – sqtrat Jan 26 '15 at 13:39
  • $\begingroup$ I didn't say it's not trivial, but you can agree the proof wasn't very precise. $\endgroup$ – user4205580 Jan 26 '15 at 13:43

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