Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X,\mathcal{M})$ be a measurable space. The definition of a simple function on a set $X$ is that it is a finite linear combination, with real coefficients, of characteristic functions of sets in $\mathcal{M}$. I'm trying to understand why, equivalently,

$f:X\rightarrow \mathbb{R}$ is simple iff $f$ is measurable and the range of $f$ is a finite subset of $\mathbb{R}$.

The forward direction makes sense by definition of what a simple function is. However, I am having trouble proving the other direction.

If $f$ is measurable, then $f^{-1}(E)\in \mathcal{M}$ for any $E\subseteq range(f)=\left\{a_1, \ldots , a_n \right\}$. What do I do from here? How do I show $f$ is simple?

I appreciate the help!

share|cite|improve this question

$f = \sum_{i=1}^{n} a_{i} \chi_{f^{-1}(\{a_i\})}$

share|cite|improve this answer

For each $a_i$, put a little neighborhood $V_i = (a_i - \epsilon, a_i + \epsilon)$ around it, so that all the neighborhoods are distinct. As $f$ is measurable,

$$A_i := f^{-1}(V_i)$$

is a measurable subset of $\mathbb{R}$. Now conclude that $$f(x) \in V_i \iff f(x) = a_i$$

and use this to show that $$f = \sum_{i = 1}^n a_i \chi_{A_i}$$

Morally: If the value of $f$ is close to $a_i$, then it had to be $a_i$ after all.

share|cite|improve this answer
Okay, I think I was stressing over a trivial thing here. Right, if $f(x)\in V_j$, then $f(x)=a_j$, so $\chi_{A_j}(x)=1$ and $\chi_{A_i}(x)=0$ for all $i\neq j$. Then we can trivially rewrite $f(x)=a_j=\displaystyle \sum_{i=1}^n a_i \chi_{A_i}$. – Sarah Jan 31 '14 at 1:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.