Is measurable function possible in discrete metric space?

Question

The function defined by $f:X{\rightarrow}Y$ is continuous. If $X$ is endowed with discrete metric. 

Now,consider characteristic function

$\chi_{A}:X{\rightarrow}R$ defined by

$\mathcal{X}_E(x) =\begin{cases}1, if& x \in E\\ 0, & if& x \not\in E \\ \end{cases}$

is continuous if $X$ is $R$ endowed with the discrete metric. 

And Now,

let (X, M) is a measurable space ,

where sigma algebra $M =\{\phi, X, \{a\}, \{b\}\}$ and $X=\{a, b\}$.

Then , the characteristic function is measurable.

Let $E\subset R$,

$\mathcal{X}_A^{-1}(E) =\begin{cases}X, & 0,1 \in E \\A, & 1 \in E, 0 \notin E \\A^C, & 1 \notin E, 0 \in E \\\emptyset, & o.w.\end{cases}$

This gives our characteristic function is measurable and continuous.  And according to definition of simple function a complex measurable function s on a measurable space $X$ whose range is only finitely many points is called simple function.  Hence,the characteristic function consider above is continuous simple function. 

Thus,giving us a continuous simple function. I am correct or wrong please check and correct my understanding .I have used Rudin book of Real and complex analysis and some online notes .

Answer 1

Yes you are correct. Since every set is open and every set is measurable, we have that every function is continuous and measurable. And a characteristic function is also simple as taking only two values.