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

Then, ball $B(x,\dfrac{1}{2})=\{x\}$ for all $x\in X $.

For any $\epsilon >0$

$f(B(x,\dfrac{1}{2}) \subset B(f(x), \epsilon) $

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 discontinuous if $X=R$ endowed with the standard metric. 

But, 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 .

  • 2
    $\begingroup$ Did you mean $X$ has the discrete metric? If so, then every function from $X$ to any topological space is continuous. $\endgroup$ – Chris Eagle Dec 3 '20 at 22:35
  • 3
    $\begingroup$ Yes, I mean X is discrete metric. What about the function being measurable. $\endgroup$ – SHREYA PANDEY Dec 4 '20 at 0:47

If a topological space $X$ is discrete then any its subset is open (and so Borel) so any function from $X$ to a topological space is continuous, and any function from $X$ to a measurable space is Borel-measurable.


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