Definition of simple function

Question

Firstly I'm sorry for this dumb question, anyway I can't grasp the real meaning of simple function, the starting point to construct the Lebesgue integral, and I hope someone will help me. Basically, I don't understand why the simple function should be written as a finite linear combination of characteristic functions. $$s(x)=\sum_{k=1}^n a_n \chi_{A_n}(x)$$ Why should we express this idea by taking a sum? I'm not an advanced student in maths, I think my problem is just understanding the meaning of this summation, so I really need a very simple answer. Moreover, I know that such a representation is not unique, anyway how could we make it unique? Many thanks.

Answer 1

It's simply superposition of each of the individual terms. Outside each set $A_k$, the term vanishes; inside this set, the term has value $a_k$. The terms therefore don't "interfere" with each other. (I'm assuming the $A_k$ are pairwise disjoint.) you obtain the simple function by layering together these even "simpler" pieces.

More: Note that you can write

$$s(x) = \begin{cases} a_1, & \textrm{ if }x\in A_1 \\ a_2, & \textrm{ if }x\in A_2 \\\ \cdots & \\ a_n, & \textrm{ if }x\in A_n \\ 0, & \textrm{ otherwise }\\ \end{cases}$$

But since the $A_k$ are presumably pairwise disjoint (if not, union them up and replace them by a single such set), we have that any $x$ will belong to at most one of them. Another way to say that is that $\chi_{A_k}(x)$ will be nonzero (since it takes the value $1$) for at most one value of $k$. Since all of the others are zero, you can add them all up (you're mostly adding up a bunch of $0$ terms with at most one nonzero term); of course, the $k$ will depend on $x$ if it exists:

$$s(x) = a_1\underbrace{\chi_{A_1}(x)}_0 + a_2\underbrace{\chi_{A_k}(x)}_0 + \cdots + a_k\underbrace{\chi_{A_k}(x)}_1 + \cdots + a_n\underbrace{\chi_{A_n}(x)}_0$$ $$\require{cancel}= \cancel{a_1\cdot 0} + \cancel{a_2\cdot 0} + \cdots + a_k\cdot 1 + \cdots + \cancel{a_n\cdot 0}$$ $$=a_k$$

If $x$ isn't in any of the $A_k$, then all of the terms are zero and $s(x)=0$, uninteresting, but still equal to that sum.