Constant Function is Measurable/Proof 2

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

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.

So, by Pointwise Limit of Measurable Functions is Measurable:

$f$ 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.

So, by Pointwise Limit of Measurable Functions is Measurable:

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

$\blacksquare$