Constant Function is Measurable/Proof 1

Proof
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 $r \in \R$.