Constant Function is Measurable/Proof 2

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.