[Math] Sheldon Axler Measure Integration Real Analysis Section 2B Exercise 11

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I apologize for the referential title, but the question is long and this book is freely available online.

I'm referring to Sheldon Axler's new book, Measure, Integration, and Real Analysis, Section 2B Exercise 11.

Specifically, given $ T $ a $ \sigma $ -algebra on $ \Gamma $ and $ X \in T $ ,

Let S = {E $\in $ T | E $ \subset $ X}

Show S = {F $ \cap $ X | F $ \in $ T}

This is one of those questions where I feel like it should be easy intuitively but then I'm just not experienced enough with sigma-algebras to make it work.

Here's how I'm reading the statement in question in my head:

Elements of a sigma-algebra that are proper subsets of an element, X in the sigma algebra, are intersections of the element X and elements of the sigma-algebra

I've attached a picture of my best attempt (since there are pictures).[![sig][1]][1]

Based on this work, I don't know if I am reading the question wrong, or if there is a typo, since the final line shows a case where X could be a subset of some other element in the sigma-algebra, and this would mean that the intersection is not a proper subset.

If it is indeed a typo, I'm extremely happy for my effort since it is not already in the errata.
If not, then maybe which properties of sigma-algebras should I be focusing on?

Thank you!

Best Answer

The question and the conclusion of it are correct.

Let $\mathscr{S}=\{E\in\mathscr{T}:E\subset X\}$ and $\mathscr{S}'=\{E\cap X: E\in\mathscr{T}\}$

  1. Show $\mathscr{S}\subset \mathscr{S}'$:

Suppose $E\in \mathscr{S}$ then $E\in\mathscr{T}$ and $E=E\cap X$ and so $E\in\mathscr{S}'$.

  1. Show that $\mathscr{S}'\subset\mathscr{S}$:

Suppose $F\in \mathscr{S}'$. Then $F=E\cap X$ for some $E\in\mathscr{T}$. Since $X\in\mathscr{T}$, then so is $F$. Finally, $F\subset X$; hence $F\in\mathscr{S}$.

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