definition of measurability of extended real value function

Question

Hi: I just started reading Bartle's "Elements of Integration" which I really like so far. It concentrates only on the lebesgue integral and leaves general measure theory for other books which I think clarifies a lot of confusion, atleast for me. Anyway, on page 10, he extends the real numbers to include $+\infty$ and -$\infty$. ( and defines positive functions and negative functions but that's not so important for my question ). So, here goes my confusion.

Assume a fixed measurable space $(X, \textbf{X})$ where $\textbf{X}$ denotes the $\sigma$ algebra on $X$. ( I don't know why he uses X twice which to me, only makes things more confusing when he's generally EXTREMELY clear. )

Then, quoting Definition 2.7 on the bottom of page 10:

An extended real valued function is $\textbf{X}$ measurable if the set $\{ x \in X: f(x) \gt \alpha \} \in \textbf{X}$ for each $\alpha$.

The collection of all extended real measurable functions on $\textbf{X}$ on $X$ is denoted as $M(X,\textbf{X})$

(The next two equalities are what I don't understand.)

Observe that if $f \in M(X, \textbf{X})$, then

\begin{equation} \{x \in X : f(x) = \infty \} = \cap_{n=1}^\infty \{x \in X : f(x) \gt n \} \end{equation}

\begin{equation} \{x \in X : f(x) = -\infty \} = \overline {\cup_{n=1}^\infty \{x \in X : f(x) \gt -n \}} \end{equation}

I don't understand either of the two equalities. Here's some reasoning behind my confusion. The set on the RHS of the first equation is monotone decreasing so it has to be equal to $ \{x \in X : f(x) \gt \infty \}$ which clearly has no meaning.

DeMorgan's law can be used to make a similar statement about the second equality. So, I'm not seeing how they can be equalities. Also, if an explanation is too long or laborious then if someone knows of a book where this is explained, that's fine also. Thanks a lot.

Note that this question is not a knock on the book. I've read ahead some and I can tell already that it's awesome. The explanations are pretty crystal clear in general except for this one Definition.

Answer 1

Fix $x \in X$. If $x \in \lbrace x \in X : f(x) = \infty \rbrace$, then $f(x) > n$ for all $n \in \mathbb{N}$, so $x \in \lbrace x \in X : f(x) > n \rbrace$ for all $n \in \mathbb{N}$. This gives us that $x \in \bigcap_{n=1}^{\infty} \lbrace x \in X : f(x) > n \rbrace$, and we get the inclusion $$\lbrace x \in X : f(x) = \infty \rbrace \subset \bigcap_{n=1}^{\infty} \lbrace x \in X : f(x) > n \rbrace.$$ Now, if $x \notin \lbrace x \in X : f(x) = \infty \rbrace$, then $x \in \overline{\lbrace x \in X : f(x) = \infty \rbrace} = \lbrace x \in X : f(x) < \infty \rbrace$. Then $f(x) < n_{0}$ for some $n_{0} \in \mathbb{N}$, and so $x \notin \lbrace x \in X : f(x) > n_{0} \rbrace$, which means that $x \notin \bigcap_{n=1}^{\infty} \lbrace x \in X : f(x) > n \rbrace$. By contrapositive, we get the reverse inclusion $$\lbrace x \in X : f(x) = \infty \rbrace \supset \bigcap_{n=1}^{\infty} \lbrace x \in X : f(x) > n \rbrace.$$ So, we get that your first equation holds. Similar logic for the argument to show the second one holds. Admittedly, I'm just now seeing probably why you're confused. The notation $\bigcup_{n=1}^{\infty}$ is usually synonymous with $\bigcup_{n \in \mathbb{N}}$, so I don't think the author meant to include $\lbrace x \in X : f(x) > \infty \rbrace$ in the intersection (or else, yeah, you'd just be getting the empty set).

Answer 2

Note that $\alpha$ is real (not an infinity) in the set $M_\alpha=\{ x | f(x) > \alpha \}$ and $\alpha < \infty$ for all real $\alpha$.

Hence if $x$ is such that $f(x) = \infty$, then $x \in M_\alpha$ for all $\alpha$.

If $x$ is such that $f(x)$ is real then there is some real $\alpha$ (take $\alpha = f(x)+1$, for example) such that $x \notin M_\alpha$.

Hence $\{x | f(x) = \infty \} = \cap_{\alpha \text{ real}} M_\alpha$.

For the other inequality, note that $\cup_{\alpha \text{ real}} M_\alpha = \{ x | f(x) \in \mathbb{R} \} \cup \{x | f(x) = \infty \}$.

Hence $(\cup_{\alpha \text{ real}} M_\alpha)^c = \{x | f(x) = -\infty \}$.

(Note that the usual $\sigma$-algebra needs to be augmented to include sets that include the infinities.)