Why Are Constant Functions Measurable?

A constant function does not need any special properties of the domain or any particular measure to be measurable. The key is that its inverse images can only be two simple sets: the whole domain or the empty set. Both are included in every sigma-algebra.

Measurability in terms of inverse images

Let $(X,\Sigma)$ be a measurable space, where $X$ is a set and $\Sigma$ is a sigma-algebra of subsets of $X$. A function $f:X\to\mathbb{R}$ is measurable if, for every Borel set $B\subseteq\mathbb{R}$, the inverse image

$$f^{-1}(B)=\{x\in X:f(x)\in B\}$$

belongs to $\Sigma$.

This definition checks which inputs are sent into a chosen set $B$. It does not require the function’s individual values, or the sets $B$ themselves, to be elements of $\Sigma$; $\Sigma$ contains subsets of the domain.

Proof for a constant function

Suppose $f(x)=c$ for every $x\in X$, for some fixed $c\in\mathbb{R}$. Choose any Borel set $B\subseteq\mathbb{R}$. There are two possibilities:

• If $c\in B$, then every input maps into $B$, so $f^{-1}(B)=X$.
• If $c\notin B$, then no input maps into $B$, so $f^{-1}(B)=\varnothing$.

A sigma-algebra always contains both $X$ and $\varnothing$. Therefore $f^{-1}(B)\in\Sigma$ for every Borel set $B$, and $f$ is measurable.

The same argument works if the codomain is any measurable space $(Y,\mathcal{T})$: a constant map $f(x)=y_0$ has preimage $X$ or $\varnothing$ for every set in $\mathcal{T}$. Thus every constant map between measurable spaces is measurable.

Worked example

Take $X=[0,1]$ with the Lebesgue sigma-algebra, and define $f(x)=\pi$. Consider the Borel set $B=(0,4)$. Since $\pi\in(0,4)$, every point of $X$ maps into $B$, and

$$f^{-1}((0,4))=[0,1].$$

For the Borel set $C=[4,\infty)$, we have $\pi\notin C$, so

$$f^{-1}([4,\infty))=\varnothing.$$

Any other Borel set gives one of these two types of preimage. Both are Lebesgue measurable, so the constant function is measurable. The measure of the domain is irrelevant to this proof; measurability depends on the sigma-algebras, not on the numerical values assigned by a measure.

Equivalent test for real-valued functions

A common equivalent criterion is to check that the sets

$$\{x\in X:f(x)>a\}$$

are in $\Sigma$ for every real number $a$. For $f(x)=c$, this set is $X$ when $c>a$ and $\varnothing$ when $c\leq a$. So the same whole-domain-or-empty-set argument proves measurability under this criterion too.

Common mistakes

• Confusing the value with a measurable set: Measurability does not ask whether $c$ is measurable. It asks whether inverse images of measurable codomain sets belong to the domain sigma-algebra.
• Thinking a measure is needed for the proof: A measure assigns sizes to sets, but the definition of a measurable function uses only the sigma-algebras. No finiteness assumption on the measure is needed.
• Confusing measurability with integrability: A measurable constant need not be integrable on a space of infinite measure. For instance, if $c\ne 0$ and $\mu(X)=\infty$, then $\int_X |c|\,d\mu=\infty$. That does not affect measurability.
• Forgetting the domain structure: The statement assumes the domain has a sigma-algebra. Without one, measurability is not defined.