Let $\mathcal{B}$ denote the Borel sigma algebra. Is it true that
$\mathcal{B}([0,1]) = \{A \cap [0,1] \ | \ A \in \mathcal{B}(\mathbb{R})$} ?
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Using the distributive property for sets and the definition of $\mathcal{B}(\mathbb{R})$, the left side can be written ${\bigcup_i (A_i \cap [0,1]) | A_i=(a,b), a,b \in \mathbb{R} }$. Each $(A_i \cap [0,1]) \in \mathcal{B}([0,1]$, which proves inclusion of the left side in the right side. The other direction is a straightforward application of definitions. – Hinrik Ingolfsson Jan 30 '24 at 22:10
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@HinrikIngolfsson i don't see why you can write the left side in the way you do it. Just because the Borel sigma algebra is generated by the open intervals it doesn't mean that any borel set can be written as union of open intervals, right? – Mac Menders Jan 31 '24 at 13:22
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but this thread answers my question: https://math.stackexchange.com/questions/2693731/induced-borel-sigma-algebra?rq=1 – Mac Menders Jan 31 '24 at 13:29