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Let $X$ be a locally compact Hausdorff (l.c.H.) space with topology $\tau$. Then one usually defines $\mathcal{B}(X) = \sigma(\tau)$ as the smallest $\sigma$-algebra that contains the open sets $\tau$.

The following construction can be found in Dinculeanu, "Vector Measures", chapter III. For every compact subset $K \subseteq X$ the Borel $\sigma$-algebra $\mathcal{B}(K)$ on $K$ can be considered as a $\sigma$-ring on $X$ that is then contained in $\mathcal{B}(X)$. Set $\mathcal{B}_c(X) := \bigcup_K \mathcal{B}(K) \subseteq \mathcal{B}(X)$ where the union is taken over all compact subsets $K \subseteq X$. In general, $\mathcal{B}_c(X)$ is not a $\sigma$-algebra (it is the $\delta$-ring generated by compact subsets of $X$). Denote by $\mathcal{T}(X) := \{ E \subseteq X \mid A \cap E \in \mathcal{B}_c(X) \text{ for all } A \in \mathcal{B}_c(X) \}$. Then $\mathcal{T}(X)$ is a $\sigma$-algebra.

Dinculeanu refers to $\mathcal{T}(X)$ as the $\sigma$-algebra of Borel subsets of $X$.

Now I was confused about the relation between the "Borel $\sigma$-algebra" $\mathcal{B}(X)$ and the "$\sigma$-algebra of Borel sets" $\mathcal{T}(X)$. For a general l.c.H. space $X$ he shows that $\mathcal{B}(X) \subseteq \mathcal{T}(X)$. If $X$ is $\sigma$-compact then it is also not hard to show that $\mathcal{B}(X) = \mathcal{T}(X)$. But there are also examples of l.c.H spaces that are not $\sigma$-compact and for which this equality holds, e.g. if $X$ is an arbitrary uncountable set equipped with the discrete topology $\mathcal{P}(X)$ then $\mathcal{B}(X) = \mathcal{P}(X) = \mathcal{T}(X)$.

Does the equality $\mathcal{B}(X) = \mathcal{T}(X)$ hold for all l.c.H. spaces?

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I'm not sure this will work. Let $X=\cup_{i\in \omega_1}I_i$ be the topological sum of $I_i:i\in \omega_1$ where $i\ne j\to I_i\cap I_j=\phi$ and each $I_i$ is homeomorphic to $I=[0,1]$....Now for any space $Y$ let $b(0,Y)$ be the Boolean algebra on $Y$ generated by the open sets. For $i\in \omega_1$ let $b(i+1,Y)$ be the algebra generated by intersections of countable subsets of $b(i,Y).$ When $i$ is a non-zero limit ordinal let $b(i,Y)=\cup_{j<i}b(j,Y). $ Then $B(Y)=b(\omega_1,Y).$ It is known (proved,I think, by Borel) that $B(I)\ne b(i,I)$ for any $i\in \omega_1. $ So $\forall i\in \omega_1 :b(i+1,I_i)\backslash b(i,I_i)\ne \phi.$....Now suppose $$T_i\in b(i+1,I_i)\backslash b(i,I_i) \text { for each } i\in \omega_1.$$ $$\text { Conjecture: } \cup \{T_i :i\in \omega_1\}\in (T(X)\backslash B(X).$$

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  • $\begingroup$ Are the sets $b(i,Y)$ related to the Borel hierarchy? $\endgroup$
    – yada
    Commented Dec 10, 2015 at 20:46
  • $\begingroup$ The sets b(i,Y) are closely related to the Borel hierarchy.For example b(i,Y) contains the F-sigma and G-delta sets but if Y is a subspace of R with int (Y) non-empty then b(1,Y) does not contain all $F_{\sigma \delta}$ sets. If we let the Borel hierarchy be $\{B(j,Y) : j \in \omega_1\}$ then $\forall i\in \omega \exists j\in \omega_1 (b(i,Y\subset B(j,Y))$. $\endgroup$ Commented Dec 11, 2015 at 6:00

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