Measurability of extended real-valued functions

measurable-functionsmeasure-theoryreal-analysis

Let $(X,\mathcal{M})$ be a measure space and consider the extended real numbers $\bar{\mathbb{R}} = \mathbb{R} \cup \{\infty,-\infty\}$ and its $\sigma$-algebra $\mathcal{B}_{\bar{\mathbb{R}}} = \{E \subset \bar{\mathbb{R}} | E \cap \mathbb{R} \in \mathcal{B}_{\mathbb{R}}\}$, where $\mathcal{B}_{\mathbb{R}}$ is the Borel $\sigma$-algebra of $\mathbb{R}$.

According to Exercise 1 in Chapter 2 of Folland's "Real Analysis: Modern Applications and Their Techniques", $f: \mathcal{M} \rightarrow \bar{\mathbb{R}}$ is measurable if and only if $f^{-1}(\{\infty\}), f^{-1}(\{-\infty\}) \in
\mathcal{M}$
and $f$ is measurable on $f^{-1}(\mathbb{R})$.

In Chapter 2, Folland defines the term $f$ is measurable on $A \in \mathcal{M}$ to mean that $f\upharpoonright_{A}$ is measurable on $\mathcal{M}_A = \{E \cap A | E \in \mathcal{M}\}$, where $f\upharpoonright_{A}$ is the restriction of $f$ to $A$.

So to show that $f$ is measurable on $f^{-1}(\mathbb{R})$, we can show that $f\upharpoonright_{f^{-1}(\mathbb{R})}: f^{-1}(\mathbb{R}) \rightarrow \mathbb{R}$ is measurable on $\mathcal{M}_{f^{-1}(\mathbb{R})} = \{E \cap f^{-1}(\mathbb{R}) | E \in \mathcal{M}\}$.

But $f\upharpoonright_{f^{-1}(\mathbb{R})}$ is a real-valued function, so we can apply standard measurability tests, such as checking that $\left(f\upharpoonright_{f^{-1}(\mathbb{R})}\right)^{-1}((a,\infty)) \in \mathcal{M}_{f^{-1}(\mathbb{R})}$ for all $a \in \mathbb{R}$. But $\infty \notin (a,\infty)$ and $-\infty \notin (a,\infty)$, so in fact, $E = \left(f\upharpoonright_{f^{-1}(\mathbb{R})}\right)^{-1}((a,\infty)) = f^{-1}((a,\infty))$ and $E = f^{-1}((a,\infty)) \subset f^{-1}(\mathbb{R})$ so $E \cap f^{-1}(\mathbb{R}) = E$.

From this reasoning, I seem to have come to the conclusion that:

$f: \mathcal{M} \rightarrow \bar{\mathbb{R}}$ is measurable if and only if $f^{-1}(\{\infty\}), f^{-1}(\{-\infty\}) \in
\mathcal{M}$
and $f^{-1}((a,\infty)) \in \mathcal{M}$ for all $a \in \mathbb{R}$.

Is this reasoning correct?

Best Answer

Yes the reasoning is correct.

Also you can deduce and use in general that:

The function $f:X \to \bar{\Bbb{R}}$ is $\mathcal{M}-$ measurable if and only if $f^{-1}((a,+\infty]) \in \mathcal{M},\forall a \in \Bbb{R}$

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