Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having a problem in understanding clearly what simple function actually means . Royden says: A real-valued function $\phi$ is called simple if it is measurable and assumes only a finite number of values. If $\phi$ is simple and has the ${\alpha_1,\alpha_2,.....\alpha_n}$ values then $\phi=\sum_{i=1} ^n \alpha_i\chi_{A_i}$, where $A_i=${x:$\phi$(x)=$\alpha_i$}.

first question is : does $\phi$ have measure zero ? ( because it has a finite number of elements)

Why do we write the simple function in such a linear combination ?

Suppose if i have to write a general function in terms of a simple function, what should i take care of , so that a function can be written in terms of simple function?

Is there a geometric presentation to understand this concept ?

Thanks for the help!

share|improve this question
1  
What do you mean $\phi$ have measure 0? It's a function! You can imagine you have some sets $A_i$, pairwise disjoint. in $A_i$, f takes value $\alpha_i$. You want to write it as a linear combination, because it's very easy to calculate the integral of these simple functions. Then you prove for any non-negative measurable function f, f can be approximated as a pointwise limit of simple functions. From that, you can define the integral of measurable function. –  user1412 May 16 '12 at 10:05
    
@LongMai, Although it is not directly relevant to the question, the concept of Radon measure is a measure for functions. Bourbaki actually develops measure theory for functions, not sets directly (you pass by indicator functions to treat sets). –  plm May 16 '12 at 10:21
    
Really? it sounds very surprising, and totally open-minded to me. I should read more about it. I'm curious what does the measure of a function tell about the function? –  user1412 May 16 '12 at 10:31
    
@LongMai, Wiki actually defines real valued Radon measures as functionals, and they conflate a little the measure on functions and the corresponding measure on sets. This is natural but may be confusing in some phrases. en.wikipedia.org/wiki/Radon_measure –  plm May 16 '12 at 11:10

3 Answers 3

up vote 6 down vote accepted

Since $\phi$ is a function (and not a subset of the measure space) we can't really speak of its measure.

Simple functions are sort of "step functions" in the following sense; the sets $\{A_{i}\}_{i=1}^{n}$ form a partition of the measure space $X$ and $\phi$ takes a constant value in each $A_{i}$, i.e. $\phi|_{A_{i}}=\alpha_{i}$ for all $i$. The most economical way of writing this is through the indicator functions of each $A_{i}$, for example by setting $\phi=\sum_{i=1}^{n}\alpha_{i}\chi_{A_{i}}$.

A simple example of a simple-function is the indicator function $\chi_{A}$ of any $A\subset X$: it takes the value $1$ in $A$ and $0$ in the complement $A^{c}$.

The significance of simple-functions is that the measure-integral is defined through them. In fact, one can show that for any non-negative measurable function $f$ there exists a nondecreasing sequence of simple functions $(\phi_{i})$ so that $\phi_{i}\to f$ point-wise.

share|improve this answer

A simple function on $X$ is constant on a finite number of measurable subsets covering $X$.

I think you are mistaken asking, in the present context, for the measure of $\phi$ a function. $\phi$ itself does not have a finite number of elements. There are ways to measure functions but they are not relevant here.

The way to picture a simple function is as a stair function: its graph is horizontal line segments over the $A_i$ abscissas, it as only a finite number of distinct heights one for each $A_i$, which is measurable. The finite linear combination makes this apparent.

As simple functions only take finitely many values they cannot represent other continuous functions than constants, so you have to use limiting processes. You may take all simple functions inferior or equal to a given function $f$, then the supremum of their values at $x$ is $f(x)$, and you may take sequences of such simple functions to approximate $f$. This is used to define the integral of $f$ and it is consistent with Lebesgue's dominated convergence theorem.

share|improve this answer
    
Well ok, Thomas E. was quicker than I. And really my answer is very similar but I did not copy. I just replied to the OP's question as felt natural. –  plm May 16 '12 at 10:10
    
@ plm very nice explanation. –  Theorem May 16 '12 at 10:14
1  
By the way, a simple function is continuous iff $\partial A_{i}=\emptyset$ for all $i$, so they can represent other continuous functions than constants depending on the underlying topology. In $\mathbb{R}^{n}$ and Euclidean topology you're right though, since the only sets without boundary are $\mathbb{R}^{n}$ itself and $\emptyset$. –  Thomas E. May 16 '12 at 10:14
    
@ThomasE., Very true. :) I had in mind the real line and vector spaces over $\mathbb R$. Thanks for the correction. –  plm May 16 '12 at 10:17
  1. Say you have a measure space $(X,\Sigma,\mu)$ (the point is general, but you can let $X=\mathbb{R}$, $\Sigma$ be the Borel sets, and $\mu$ be Lebesgue measure) and a meaurable function $f:X\to\mathbb{R}$. It doesn't make sense to ask whether $f$ has measure zero in this context, since the measure is only defined for elements of $\Sigma$, which are all subsets of $X$.

  2. It is not possible to write general measurable functions as simple functions. But you can write them always as the pointwise limit of simple functions. And a nonnegative measurable function can be written as the increasing limit of simple functions. This is not trivial and requires proof. I'm sure Royden shows that somewhere.

  3. The motivation for simple functions is the following: We don't have an obvious notion for what the integral of a general measurable function should be. Let us for simplicity consider the case in which the measure is Lebesgue measure and restrict the functions in question to be non-negative. The integral should be something like the area under the function. We know that the area of a rectagle is it's height times its width. Now if you let $A$ be the interval $[a,b]$ and look at the function $\alpha\chi_A$, it will look like a rectangle with height $\alpha$ and width $b-a=\lambda(A)$. If we add such functions, we can add the area, so if all the $A_i$ in the simple functions $\sum_{i=1}^n \alpha_i \chi_{A_i}$ are intervals with $A_i=[a_i,b_i]$, we can calculate the integral as $\sum_{i=1}^n\alpha_i(b_i-a_i)=\sum_{i=1}^n\alpha_i\lambda(A_i)$. Now it doesn't really matter that the $A_i$ are intervals and have a "length", we can use Lebesgue measure to generalize this costruction by allowing the $A_i$ be arbitrary measurable sets, which gives us a much larger class of functions for which it is "obvious" what the integral should be. Now if a sequence of functions for which the integral is obvious converges to a function, then the areas und these functions should converge too. So for a general nonnegative measurable function $f$, we pick a sequence of simple functions $(f_n)$ that increases to $f$ and define $\int fd\mu$ to be $\lim_{n\to\infty}\int f_n d\mu$. It can be shown that this definition works, the specific sequence of simple functions does not matter.

share|improve this answer
    
Thanks for a good explanation ! –  Theorem May 16 '12 at 10:24

Your Answer

 
discard

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.