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## Homework Statement

I was hoping that someone would be able to help me solve this problem regarding simple functions and their measure. This problem is coming straight from

**Introduction to Measure Theory**by Terrence Tao. A link to the free version is attached at the end of this post.

Show that an unsigned function ##f: \mathbb{R}^d \to [0, +\infty]## is a simple function if and only if it is measurable and takes on at most finitely many values.

## Homework Equations

The definitions that I am working with are as follows:

An

**unsigned simple function**##f: \mathbb{R}^d \to [0, +\infty]## is a finite linear combination ##f= c_11_{E_1}+ \ldots + c_k1_{E_k}## of indicator functions ##1_{E_i}## of Lebesgue measurable sets ##E_i \subset \mathbb{R}^d## for ##i=1, \ldots , k,## where ##k \geq 0##.

An unsigned simple function ##f:\mathbb{R}^d \to [0, +\infty]## is

**unsigned Lebesgue measurable**, or

**measurable**for short, if it is the point-wise limit of unsigned simple functions, i.e., if there exists a sequence ##f_1,f_2,f_3, \ldots : \mathbb{R}^d \to [0, +\infty]## of unsigned simple functions such that ##f_n(x) \to f(x)## for every ##x \in \mathbb{R}^d##.

## The Attempt at a Solution

My attempt at the solution is this:

If ##f:\mathbb{R}^d \to [0, +\infty]## is simple, then ##f## is a linear combination ##f=c_11_{E_1}+ \ldots + c_n1_{E_n}## of indicator functions ##1_{E_i}## of Lebesgue measurable sets ##E_i \subset \mathbb{R}^d## for ##i =1, \ldots, n##, where ##n \in \mathbb{N}## and ##c_1, \ldots, c_n \in [0, +\infty]##. We wish to show that this implies that ##f## is measurable and takes on at most finitely many values. To say that it is measurable would mean that ##\exists f_1, f_2, \ldots## each unsigned simple such that ##\forall x \in \mathbb{R}^d, f(x) = \displaystyle \lim_{n \to \infty} f_n(x)##. If we cut up ##[0, +\infty]## into finitely many boxes ##B##, each small enough that ##\forall x, y \in B, |f(x)-f(y)| < \frac{1}{n}##, or ##f(x), f(y) \in [0, +\infty].## ##f_n(x) = \inf(f(B))## if ##x \in## box ##B##. ##f_n(x)=0## for ##x \notin [0, +\infty]##.

Then ##\forall x \in \mathbb{R}^d##,

##f_n(x) \to f(x)## and ##f_n = \displaystyle \sum_{\text{boxes B}}\inf(f(B))1_B## is simple.

To prove the backwards direction, if ##f## is measurable, then ##\exists f_1,f_2 \ldots## each unsigned simple such that ##\forall x\in \mathbb{R}^d, f(x)=\displaystyle \lim_{n=\infty}f_n(x)##. Since ##f## is composed of simple functions converging point-wise, the result follows by definition.

Note: In this proof, I am imitating an approach we used in class to show that if a function is continuous, then it is measurable. I am not sure if it still applies in this case, which is why I would really appreciate some feedback in order to help me solve this problem.

Thank you very much in advance.

Link to the textbook:

http://terrytao.files.wordpress.com/2011/01/measure-book1.pdf