Show that $\sigma(\mathcal{F}^d)\subseteq \mathcal{B}(\mathbb{R}^d)$

Question

Let $\mathcal{F}^d$ be the collection of sets that can be written as the finite union of half open rectangles $(a,b]^d$.

Exercise: Show that $\sigma(\mathcal{F}^d)\subseteq \mathcal{B}(\mathbb{R}^d)$.

My Question:

In the solution to this problem it is stated that it is enough to show that $\mathcal{F}^d\subseteq\mathcal B( \mathbb{R}^d)$.

Why is that true?

I know that $\sigma(\mathcal{F}^d)\subseteq\mathcal{P}(\mathcal{F}^d)$, but I don't see why $\mathcal{F}^d\subseteq \mathcal{B}(\mathbb{R}^d)\Rightarrow \sigma(\mathcal{F}^d)\subseteq \mathcal{B}(\mathbb{R}^d)$.

Thanks!

Answer 1

I suspect that $\mathcal B(\mathbb R^d)$ denotes the Borel $\sigma$-algebra on $\mathbb R^d$

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Let $\Omega$ be a set and let $\mathcal V$ and $\mathcal A$ be subcollections of $\wp(\Omega)$.

It can be shown that the intersection of all $\sigma$-algebras on $\Omega$ that contain $\mathcal V$ is itself a $\sigma$-algebra that contains $\mathcal V$.

It is denoted by $\sigma(\mathcal V)$.

So if $\mathcal V\subseteq\mathcal A$ where $\mathcal A$ denotes a $\sigma$-algebra then by definition $\sigma(\mathcal V)\subseteq\mathcal A$.

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This can be applied on $\Omega=\mathbb R^d$, $\mathcal V=\mathcal F^d$ and $\mathcal A=\mathcal B(\mathbb R^d)$