# Constant Function is Measurable

## Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f : X \to \overline \R$ be a constant extended real-valued function.

That is, there exists $c \in \overline \R$ such that:

$\map f x = c$ for all $x \in X$.

Then $f$ is $\Sigma$-measurable.

## Proof 1

By the definition of a $\Sigma$-measurable function, we aim to show that:

$\set {x \in X : \map f x \le r}$ is $\Sigma$-measurable for each $r \in \R$.

First suppose that $\size c < \infty$.

Note that there are no $x \in X$ such that $\map f x < c$.

So for $r < c$, we have:

$\set {x \in X : \map f x \le r} = \O$

From $\sigma$-Algebra Contains Empty Set, we then have:

$\set {x \in X : \map f x \le r} \in \Sigma$ if $r < c$.

Since for all $x \in X$ we have $\map f x = c$, we also have $\map f x \le c$ for all $x \in X$.

So, if $r \ge c$:

$\set {x \in X : \map f x \le r} = X$

From the definition of a $\sigma$-algebra, we then have:

$\set {x \in X : \map f x \le r} \in \Sigma$ if $r \ge c$.

So:

$\set {x \in X : \map f x \le r} \in \Sigma$ for all $r \in \R$

which was the demand.

Now suppose that $c = \infty$.

Then we have:

$\map f x \ge r$

for all $x \in X$ and $r \in \R$.

So:

$\set {x \in X : \map f x \le r} = \O$

for each $r \in \R$.

From $\sigma$-Algebra Contains Empty Set we have $\O \in \Sigma$, so:

$\set {x \in X : \map f x \le r} \in \Sigma$ for all $r \in \R$.

Now suppose that $c = -\infty$.

Then we have:

$\map f x \le r$

for all $x \in X$ and $r \in \R$.

So:

$\set {x \in X : \map f x \le r} = X$

for each $r \in \R$.

From the definition of a $\sigma$-algebra, we have that $X \in \Sigma$, so:

$\set {x \in X : \map f x \le r} \in \Sigma$ for all $r \in \R$.

$\blacksquare$

## Proof 2

First, suppose that $\size c < \infty$.

From Characteristic Function of Universe, we can write:

$\map f x = c \map {\chi_X} x$

for each $x \in X$.

From the definition of a $\sigma$-algebra, we have:

$X \in \Sigma$

So:

$f$ is a simple function.

Then, from Simple Function is Measurable, we have:

$f$ is $\Sigma$-measurable.

Now suppose that $c = \infty$.

For each $n \in \N$, define $f_n : X \to \R$ by:

$\map {f_n} x = n$

for each $x \in X$.

Then:

$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$

for each $x \in X$.

We have shown that each $f_n$ is $\Sigma$-measurable for each $n \in \N$, so this is the limit of $\Sigma$-measurable functions.

$f$ is $\Sigma$-measurable in this case.

Now suppose that $c = -\infty$.

For each $n \in \N$, define $g_n : X \to \R$ by:

$\map {g_n} x = -n$

for each $x \in X$.

Then:

$\ds \map f x = \lim_{n \mathop \to \infty} \map {g_n} x$

for each $x \in X$.

We have shown that each $g_n$ is $\Sigma$-measurable for each $n \in \N$, so this is the limit of $\Sigma$-measurable functions.

$f$ is $\Sigma$-measurable in this case.

$\blacksquare$