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I am studying measure theory and I was curious about what the borel set of $Y=[0, \infty)$, $B_Y$ looks like compared to the borel set $B$ for real numbers.

Does every element in $B_Y$ take the form $Y \cap E$, with $E \in B$ ? If so could someone please show me how to prove such a thing?

Thank you.

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3 Answers 3

up vote 2 down vote accepted

Yes. If $X$ is an arbitrary topological space space and $Y \subseteq X$ is a subspace then the Borel $\sigma$-algebra satisfy $$\mathcal{B}(Y) = \lbrace E \cap Y : E \in \mathcal{B}(X)\rbrace.$$ To see this, notice that the right hand side $\mathcal{A}$ is a $\sigma$-algebra on $Y$ containing all open sets, so $\mathcal{A} \supseteq \mathcal{B}(Y)$. On the other hand, $\Sigma = \lbrace F \subseteq X : F \cap Y \in \mathcal{B}(Y)\rbrace$ is a $\sigma$-algebra on $X$ containing all open subsets of $X$, so $\Sigma \supseteq \mathcal{B}(X)$, and hence $\mathcal{B}(Y)$ contains $\mathcal{A}$.

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I don't see how to comment on the other answers but I don't understand Arthur's answer: 1. In what sense is it involved to prove that the subsets of $Y$ which are open are of the form $Y \cap U$ with $U$ open in $X$ (that's the definition of the subspace topology) 2. Where is it used that $Y$ is Borel (in fact, closed)? –  Daniel M. Dec 11 '12 at 22:32
    
Thank you for your answer! –  J Kasahara Dec 12 '12 at 17:43

It is a general fact that if you have an inclusion of topological spaces $X\subseteq Y$, then the Borel algebra of $X$ consists of traces of the Borel algebra of $Y$.

This can be shown using the fact that the open sets in $X$ are traces of open sets in $Y$, and one of the standard methods of showing that some family is the entire $\sigma$-algebra generated by open sets, the most common being:

  1. By transfinite induction over the Borel hierachy.
  2. Or, a bit simpler perhaps, by showing that the family of traces of Borel sets in $Y$ is a $\sigma$-algebra of subsets of $X$ and that it is a subfamily of the Borel algebra of $X$.
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Thank you for your answer! –  J Kasahara Dec 12 '12 at 17:41

Actually, since your set $Y$ is itself a Borel (in fact, closed) subset of $\mathbb{R}$, it turns out at the Borel subsets of $Y$ are exactly the Borel subsets of $\mathbb{R}$ which happen to be subsets of $Y$.

In general, if $Y$ is an arbitrary subspace of some topological space $X$, the Borel subsets of $Y$ are of the form $Y \cap E$ where $E$ is a Borel subset of $X$. The reason for this is because the open subsets of $Y$ are exactly those sets of the form $Y \cap U$ where $U$ is an open subset of $X$ (though the proof is somewhat more involved). If $Y$ is a Borel subset of $X$, then $Y \cap E$ is also a Borel subset of $X$ for all Borel $E \subseteq X$, which helps gives the characterisation from above.

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Thank you for your answer! –  J Kasahara Dec 12 '12 at 17:42

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