I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 11 on p.38 in Exercises 2B in this book.
Exercise 11:
Suppose $\mathcal{T}$ is a $\sigma$-algebra on a set $Y$ and $X\in\mathcal{T}$. Let $\mathcal{S}=\{E\in\mathcal{T}: E\subset X\}.$
(a) Show that $\mathcal{S}=\{F\cap X:F\in\mathcal{T}\}.$
(b) Show that $\mathcal{S}$ is a $\sigma$-algebra on $X$.
A proposition in this book:
2.25 $\sigma$-algebras are closed under countable intersection
Suppose $\mathcal{S}$ is a $\sigma$-algebra on a set $X$. Then
(c) if $E_1,E_2,\dots$ is a sequence of elements of $\mathcal{S}$, then $\bigcap_{k=1}^\infty E_k\in\mathcal{S}.$
My solution:
(a)
Let $E\in\mathcal{S}$.
Since $E\subset X$, $E=E\cap X$.
And $E\in\mathcal{T}$.
So, $E\in\{F\cap X:F\in\mathcal{T}\}$.
Conversely, let $E\in\{F\cap X:F\in\mathcal{T}\}.$
Then we can write $E=F\cap X$ for some $F\in\mathcal{T}$.
Since, $F,X\in\mathcal{T}$, $E=F\cap X\in\mathcal{T}$ by 2.25(c) in the book.
And $E=F\cap X\subset X$.
So, $E\in\mathcal{S}.$
(b)
$\emptyset\in\mathcal{T}$ and $\emptyset\subset X$.
So, $\emptyset\in\mathcal{S}$.
Let $E\in\mathcal{S}$.
Since $E\in\mathcal{T}$, $Y\setminus E\in\mathcal{T}$.
So, by 2.25(c) in the book, $X\setminus E=(Y\setminus E)\cap X\in\mathcal{T}$.
And $X\setminus E\subset X$.
So, $X\setminus E\in\mathcal{S}$.
Let $E_1,E_2,\dots$ be a sequence of elements of $\mathcal{S}$.
Then, since each $E_i$ is an element of $\mathcal{T}$, $E_1\cup E_2\cup\dots\in\mathcal{T}$.
Since each $E_i$ is a subset of $X$, $E_1\cup E_2\cup\dots\subset X$.
So, $E_1\cup E_2\cup\dots\in\mathcal{S}$.
I solved the exercise above.
But I am worried if my answer is correct because it was easy.
And I wonder why the author wrote (a).
I didn't use (a) to solve (b).
Best Answer
Your work looks fine. In my experience it is more common to define $\mathcal S$ as in part (a), but as you've seen this is equivalent to the primary definition given. This is also called the trace $\sigma$-field. More generally it is not even necessary that $X$ be measurable.