# Bounded, measurable and supported on a set of finite measure function

Suppose f is a bounded and measurable function on R and supported on a set of finite measure. Prove that for every $\epsilon \gt 0$ there exists a simple function $s$ such that $\int |f-s|dx$ $\lt \epsilon$

I have question: Can I let this simple function be supported on a set of finite measure? Then this simple function will be bounded and measurable and supported, then I can use the linearity of integral and the definition of the integral of this kind of functions to prove that.

I am wondering if I am right. I really appreciate your help if you would like to give me another proof.

Thanks

You can choose any simple function that satisfies the condition whether it is supported by finite set or not. In fact, simple function is always bounded and measurable without any condition.

• Can you give me another proof of this question, Wit@Worawit Tepsan – Yang Dec 8 '13 at 20:10

We assume that $f$ is supported on $E$ which $m(E)=L<\infty.$ So we can consider a function $f$ on E. Since $f$ is bounded, it must have a and b such that $a \leq f(x) \leq b$ for all $x\in E$. Let $N > \frac{L(b-a)}{\varepsilon}.$

We define $E_i = \{x \in E : a+i(b-a)/N \leq f(x) < a+(i+1)(b-a)/N\}$.

Choose $x_i\in E_i$ and define
$s(x) = \sum_{i=0}^{N} f(x_i) \lambda(E_i)$.

Then you can show that

$|f(x)-s(x)|\leq \frac{\varepsilon}{L}.$

And so on.

• the question is that the domain is R which has infinity measure@Worawit Tepsan – Yang Dec 8 '13 at 21:14
• I edited the proof. – Worawit Tepsan Dec 8 '13 at 21:41
• @Worawit Tepsan, can you please elaborate on how you can show that $|f(x) - s(x)|\leq \frac{\epsilon}{L}$? – ALannister Nov 18 '15 at 15:17
• I did typo. I should choose $N>\frac{L(b-a)}{\varepsilon}.$ – Worawit Tepsan Nov 29 '15 at 3:33

Let $$X$$ be the support of $$f$$ and $$\mathbb{1}_X$$ its characteristic function. Now observe that $$f=f\cdot \mathbb{1}_X$$ and that $$s \cdot \mathbb{1}_X$$ approximate $$f$$.