Showing that $\mathcal{B}(\mathbb{R^n}) \subset \mathcal{M}$

by user516079   Last Updated September 12, 2019 04:20 AM

So, I'm being asked the following:

Show that if $\mathcal{M}$ is a $\sigma$-algebra on $\mathbb{R^n}$ and all the closed sets of $\mathbb{R^n}$ are contained in $\mathcal{M}$, then $\mathcal{B}(\mathbb{R^n}) \subset \mathcal{M}$. Here $\mathcal{B}(\mathbb{R^n})$ is the Borel $\sigma$-algebra on $\mathbb{R^n}$

What confuses me about the above exercise is that it seems too simply really. We can easily show that the $\mathcal{B}(\mathbb{R^n})$ is generated by the collection of all closed sets in $\mathbb{R^n}$. But, by the definition of a $\sigma$-algebra generated by a set, $\mathcal{B}(\mathbb{R^n})$ would simply be, by definition, the intersection of all $\sigma-$algebras on $\mathbb{R^n}$ that contain the closed sets. Since $\mathcal{M}$ is such a $\sigma$-algebra, the result would follow by definition...

Am I missing something here? Surely there's some nuance in the question that I'm overlooking but failing to see? Or, is this reasoning actually correct?



Answers 1


You seem to say "it is easy to show that $\mathcal B(\Bbb R^n)$ is generated by the closed subsets of $\mathbb R^n$". If this is easy, then the rest of the proof, as you have written is correct and even easier. As Charles mentions in the comments, this helps in showing that all Borel sets are Lebesgue measurable.

However, by Dynkin's $\pi-\lambda$ theorem, it turns out that the conditions on $\mathcal M$ can be weaker : indeed, $\mathcal M$ needs to only be a Dynkin system (note that closed sets already form a $\pi$-system) so the conclusion is true even if $\mathcal M$ is not a sigma-algebra, but something weaker.

See here for an application of the $\pi-\lambda$ theorem that you will in all probability see later in your course or life : Lebesgue measure is unique

астон вілла олоф мэллбэрг
астон вілла олоф мэллбэрг
September 12, 2019 04:17 AM

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