In real analysis, is constant function measurable??

Question

From my text book, it says that for a constant function $f(x)=c,x\in E, c\in\mathbb{R}\cup \{-\infty,+\infty\} $ is measurable.

The prove is saying that if $a>c$ then $\forall x\in E\rightarrow f(x)=c<a$, so $f^{-1}((a,+\infty])=\varnothing$, and $\varnothing$ is definitely an element of borel set in $E$.

On the contrary, it can be prove that $f^{-1}((a,+\infty])=E$.

Combining the above 2, we can say that it is measurable function.

However, this prove based on a condition that $E$ is the full set of its topology.

So if $E\subset X$ and $<X,T>$ is an topology, then $\varnothing, X$ has to be an open set, but $E$ does not have to be.

Thus $E$ can be out side of borel set of $X$, which actually make $f(x)=c$ not measurable??

Thus I would like to ask if I'm right and if yes, then why the text book commit the fact that constant function is measurable??

Answer 1

When people say $f : X \to Y$, by definition of this notation, $\text{dom}(f) = X$. With that being said, the definition of a measurable function $f: X \to \mathbb{R}$ has nothing to do with the topology on $X$ (you do not need any topology of $X$). All you need is a $\sigma$-algebra $\mathcal{A}$ on $X$. $f$ is measurable if for every $a \in \mathbb{R}$, $f^{-1}((a, \infty)) \in \mathcal{A}.$

What is true is that if $X$ does happen to have a topology, one natural $\sigma$-algebra is the Borel. But you can totally have others.

The book is correct. No matter what $\sigma$-algebra you have on $X$, a constant function will be measurable.

Answer 2

This has nothing to do with topology. Suppose $(\Omega_1, \mathcal{F}_1)$, $(\Omega_2, \mathcal{F}_2)$ are measurable spaces. Pick any $c \in \Omega_2$ and consider the constant function $f: \Omega_1 \to \Omega_2, x \mapsto c$. Then for any $A \subset \Omega_2$ $$ f^{-1}(A) = \emptyset \quad\text{if c $\not\in A$} $$ and $$ f^{-1}(A) = \Omega_1 \quad\text{if c $\in A$}. $$ But $\emptyset$ and $\Omega_1$ must be in $\mathcal{F}_1$ by the definition of $\sigma$-algebras. Thus, $f^{-1}(\mathcal{F_2}) \subset \mathcal{F}_1$ and $f$ is measurable.

Answer 3

For any $q\in \mathbb Q$ we have either $\{f\leqslant q\} = \emptyset$ or $\{f\leqslant q\} = \mathrm{dom} (f)$ so $f$ is measurable.