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  1. Definition of measurable space:

If $X$ is a set and $\mathcal{M} \subset \mathcal{P}(X)$(Power set of $X$) is a $\sigma$-algebra, $(X, \mathcal{M})$ is called a measurable space and the sets in $\mathcal{M}$ are called measurable sets.

  1. Definition of $(\mathcal{M}, \mathcal{N})$-measurable:

If $(X, \mathcal{M})$ and $(Y, \mathcal{N})$ are measurable spaces, a mapping $f: X->Y$ is called $(\mathcal{M}, \mathcal{N})$-measurable, or just measurable when $\mathcal{M}$ and $\mathcal{N}$ are understood, if $f^{-1}(E) \in \mathcal{M}$ for all $E \in \mathcal{N}$.

2.1 Proposition from Folland's Real Analysis Modern Techniques:

If $\mathcal{N}$ is generated by $\mathcal{E}$, then $f: X -> Y$ is $(\mathcal{M}, \mathcal{N})$-measurable if and only if $f^{-1}(E) \in \mathcal{M}$ for all $E \in \mathcal{E}$.

What I'm confusing about is if $\mathcal{N}$ does being generated by $\mathcal{E}$, or say $\mathcal{N} = \sigma(\mathcal{E})$, is there any relationship between $Y$ and $\mathcal{E}$? Should Y be necessarily a subset of $\mathcal{E}$?

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    $\begingroup$ $\mathcal{E}$ is a subset of $\mathcal{P}(Y)$, so a family of subsets of $Y$. $Y$ may or may not be a member of $\mathcal{E}$. $\endgroup$ – Daniel Fischer Aug 28 '15 at 15:11
  • $\begingroup$ @DanielFischer: $Y$ must be a subset of $\mathcal{N}$ which is generated by $\mathcal{E}$. So $\mathcal{N}$ can be produced by two different methods? emmmm, I feel it weird. $\endgroup$ – Bear and bunny Aug 28 '15 at 15:16
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    $\begingroup$ No, $Y$ must be an element of $\mathcal{N}$. In fact it is the largest (by inclusion) element of $\mathcal{N}$. $\endgroup$ – Daniel Fischer Aug 28 '15 at 15:20
  • $\begingroup$ @DanielFischer: Sorry, my mistake. The largest element, yes, make sense. $\endgroup$ – Bear and bunny Aug 28 '15 at 15:23
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    $\begingroup$ All elements of $\mathcal{E}$ are subsets of $Y$, by definition. $\endgroup$ – Daniel Fischer Aug 28 '15 at 17:20
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Conclude from Daniel Fisher's comment,

$\mathcal{E}$ is a subset of $\mathcal{P}(Y)$ (power set of $Y$). Measurable space $(Y, \mathcal{N})$ requires $\mathcal{N}$ must be a $\sigma$-algebra on $Y$ while $Y$ must contain the fullest staffs and be a element of $\mathcal{N}$. At the same time, $\mathcal{N}$ can play a role of generated $\sigma$-algebra of some set $\mathcal{E}$.

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