# Constant functions are measurable explanation

The definition of measurable functions is: Let $\Sigma$ be a sigma algebra of set $X$. Then $f:X \rightarrow \bar{\mathbb R}$ is measurable if $\{x:f(x)>a\} \in \Sigma$ for all $a \in \mathbb R$.

Let $f(x)=c$. For any $a \in \mathbb R$, the preimage $f^{-1}(a, +\infty)$ is equal to either the empty set or $X$.

I can't see how this works out the statement. I do know that the empty set and $X$ are always in $\Sigma$ but I don't know how it came to finding out what the preimage is equal to.

Also what does the real numbers with a bar mean?

• either $a<c$ or $a\ge c$. In the former case $c\in(a,\infty)$ and the preimage is $X$. In the latter case $c\not\in(a,\infty)$ and the preimage is empty. I guess $\bar{\mathbb R}$ includes $\infty$ as an element. Commented Apr 10, 2016 at 16:31

Ok lets look at it step by step

$f(x)=c_1$ Now we need to check the $\{x\in X| f(x)>c\}$

Basically this is the same as $\{x\in X| c1>c\}$ Now we debate depending on variable $c$

For $c \ge c_1$ We get $\{x\in X| c_1>c\ge c_1\}=\emptyset$ since $c_1>c_1$ cannot happen. And as we know empty set is in sigma algebra by definition.

Now for $c < c_1$ We get $\{x\in X| c_1>c \}=X$ since $c_1<c$ is exactly how we chose $c$

And since we know that empty set is in sigma algebra and we also know that if $A\in \Sigma \implies A^c \in \Sigma$ So that means that $X \in \Sigma$

Now we know that $(\forall c \in R) \space \{x\in X| f(x)>c \}\in \Sigma$ So by definition the function is measurable.

$\bar{\mathbb R} = \mathbb{R}\cup\{-\infty,+\infty\}$

• What if we had the characteristic function of set $E$, $f(x)=1$ if $x \in E$ and $f(x)=0$ iff $x \notin E$ is measurable $\iff$ $E \in \Sigma$? We could only have that if $x \in E$, $1>a$ and $x \notin E$, $0>a$ and if those inequalities don't hold, then we have the empty set. If they hold, then we have $X$. But I don't see how we get $E$. Commented Apr 10, 2016 at 19:05
• *if ${}$ ${}$ ${}$ ${}$ Commented Apr 10, 2016 at 19:16
• For characteristic function if f(x)>c if $c \in [0,1)$ then you get E if $c \in (-\infty,0)$ you get X and for $c \in [1,+\infty)$ you get empty set Commented Apr 10, 2016 at 21:48
• Also worth noting is that characteristic function is not a constant function. But the equivalence you have put should be $\chi_E$ measurable iff $E$ measurable Commented Apr 10, 2016 at 21:52

Let $$X$$ and $$Y$$ be measurable sets with $$\Sigma$$-algebras $$\Sigma_{X}$$ and $$\Sigma_{Y}$$, respectively.

Let $$f:X\longrightarrow Y$$ be constant s.t. $$\forall x\in X: f(x)=a$$. Then $$\forall U\in\Sigma_{Y}$$ we have: $$\begin{equation*} f^{-1}(U)=\left\{\begin{matrix}X & \text{if a\in U}\\ \emptyset & \text{else.}\end{matrix}\right. \end{equation*}$$ Both $$X$$ and $$\emptyset$$ are in $$\Sigma_{X}$$.

Same argument shows that $$f$$ is continuous by the way, if you have topologies on $$X$$ and $$Y$$.