Extended real line sigma algebra

Question

Like in Folland, if we define: $\mathscr{B}_{\overline{\mathbb{R}}} = \{E \subset \overline{\mathbb{R}}: E \cap \mathbb{R}\in \mathscr{B}_{\mathbb{R}} \}$. How do we prove $\mathscr{B}_{\overline{\mathbb{R}}}= \sigma(\mathscr{C})$ where $\mathscr{C} = \{(\alpha, \infty] ; \alpha \in \mathbb{R}\}$ directly?

We didn't define topology in $\mathscr{B}_{\overline{\mathbb{R}}}$.

Answer 1

Here's a general trick to show equality of certain structures. It works for $\sigma$-algebras, topologies, subgroups of a given group, etc.

The idea is that you take a generating set $G_A, G_B$ of each object $A$ and $B$, and show that every element of the generating set $G_A$ is generated by $G_B$ and vice versa. This way you are essentially proving that the object generated by $G_A$ is contained in the object generated by $G_B$ and vice versa, making those generated objects equal. But those generated objects are exactly $A$ and $B$, making those equal.

In this case, $\sigma(\mathcal C)$ is by definition generated by $\mathcal C$. And $\mathcal B_{\overline{\mathbb R}}$ is generated by itself (I assume it is known that $\mathcal B_{\overline{\mathbb R}}$ is a $\sigma$-algebra). So we need to show that:

• every element of $\mathcal C$ is generated by elements of $\mathcal B_{\overline{\mathbb R}}$, and
• every element of $\mathcal B_{\overline{\mathbb R}}$ is generated by elements of $\mathcal C$

$\mathcal C$ generated by $\mathcal B_{\overline{\mathbb R}}$:

This is easy: It's trivially true since $\mathcal C\subset\mathcal B_{\overline{\mathbb R}}$. This is because $(\alpha,\infty]\cap\mathbb R=(\alpha,\infty)\in\mathcal B_{\mathbb R}$.

$\mathcal B_{\overline{\mathbb R}}$ generated by $\mathcal C$

We can generate every open subset of $\mathbb R$. To prove this, it suffices to show that we can generate $(a,b)$ for all $a,b\in\mathbb R$, because every open subset of $\mathbb R$ is a countable union of such intervals.

We have :

• $(a,\infty]\in\mathcal C$ by definition
• $[-\infty,b)=\bigcup_{n=1}^\infty [-\infty,b-\frac1n]=\bigcup_{n=1}^\infty(b-\frac1n,\infty]^c$
• $(a,b)=(a,\infty]\cap[-\infty,b)$

Now since we can generate every open subset of $\mathbb R$, we can also generate every element of $\mathcal B_{\mathbb R}$, since this is by definition generated by these open sets. And the elements of $\mathcal B_{\overline{\mathbb R}}$ are exactly those of $\mathcal B_{\mathbb R}$, but possibly also containing $-\infty,\infty$ or both. So if we can generate $\{\infty\}$ and $\{-\infty\}$, we are done. This is how we get those singletons:

• $\{\infty\}=\bigcap_{n=0}^\infty(n,\infty]$
• $\{-\infty\}=\bigcap_{n=0}^\infty[-\infty,-n]=\bigcap_{n=0}^\infty(-n,\infty]^c$