Measuring Unsigned Simple Functions

[jamilmalik]

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

 

[mighty2000]

Thank you for posting this interesting problem. Your attempt at the solution is on the right track, but there are a few areas that need clarification and further explanation.

Firstly, when you say that the function f is a linear combination of indicator functions, it is important to clarify that the sets E_i are disjoint, otherwise the function may not be simple. This is because if E_i and E_j are not disjoint, then the indicator functions 1_{E_i} and 1_{E_j} will overlap and their linear combination may not be a simple function.

Secondly, when you say that f_n(x) = inf(f(B)) if x is in the box B, it is important to note that this only holds if the box B is contained in the set [0, +∞]. Otherwise, f_n(x) should be set to 0.

Additionally, when you say that f_n(x) = 0 if x is not in [0, +∞], this is not entirely accurate. The function f_n(x) should be set to 0 if x is not in any of the boxes B, not just if it is not in the set [0, +∞].

Furthermore, when you say that f_n(x) converges point-wise to f(x), it is important to clarify that this is true for all x in [0, +∞]. This is because f_n(x) is only defined for x in [0, +∞], so it does not make sense to talk about its convergence for x outside of this set.

Finally, in the last part of your proof, you mention that the result follows by definition, but it would be helpful to explain why this is the case. In particular, you should explain how the fact that f_n(x) converges point-wise to f(x) implies that f takes on at most finitely many values.

Overall, your approach to the problem is correct, but it needs some clarification and further explanation. I hope this helps and good luck with your studies!