What does a simple function actually mean? 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! 
 A: *

*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 $X$.

*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.

*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.
A: 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.
A: 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.
A: I only explain that $\phi(X)$, not $\phi$, has (Lebesgue) measure zero. See the other answers for your other questions.
The range of the function $\phi:X\to\mathbb{R}$ is denoted by $\phi(X)={\alpha_1,\alpha_2,.....\alpha_n}$. We have $m(\phi(X))=0$, where $m$ is the Lebesgue measure, i.e., the set $\phi(X)$ has (Lebesgue) measure zero.
You can only measure sets, not functions (like $\phi$). To be exact, technically you could interpret a function $\phi:X\to\mathbb{R}$ as the subset $\{(x,y)\in X\times\mathbb{R}\,:\, \phi(x)=y\}$ of $X\times \mathbb{R}$ and define a measure on $X\times \mathbb{R}$, but surely that is not what you wanted to do.
I assumed above that you only think of the Lebesgue measure $m$. One could define also other measures on $\mathbb{R}$, such has the counting measure $\sigma$ (the number of elements in a set), which has $\sigma({\alpha_1,\alpha_2,.....\alpha_n})=n$ (if $\alpha_i\ne\alpha_j$ whenever $i\ne j$). Thus, $\sigma(\phi(X))=n\ne0$, so $\phi(X)$ is not $\sigma$-zero measurable, though it is $m$-zero measurable.
If, instead, $\phi:X\to\mathbb{R}^k$ or $\phi:X\to\mathbb{C}^k$ or the like, the above still holds (for the corresponding $n$ or $2n$ dimensional Lebesgue measure). In fact, for any function $\phi:X\to Y$ as long as, on $Y$, you use a measure $\mu$ for which $\mu(Q)=0$ for finite $Q$.
