Why is the Lebesgue integration definition for nonnegative functions well defined?

Question

I am following "Lebesgue Integration on Euclidean Space" by Frank Jones. In Page 121, chapter 6: Integration, he starts of by defining for $s∈S$ where $S$ is the class of measurable simple functions $s$ on $\mathbb R^n$, and $s$ is presented in the form: $$s=\sum_{k=1}^ma_k\chi A_k$$

where $0<a_k<∞$ and the sets $A_k$ are measurable and disjoint, then: $$\int sd\lambda =\sum_{k=1}^ma_k\lambda (A_k)$$

A couple of questions here:

• is it $\chi$ or $x$? If it is $\chi$ then I have no idea what it represents and if it is $x$ it is probably a typo since it looks nothing the other $x$ in the whole book.
• Why is it well defined? To my understanding to prove that it is well defined I need to show that it is always possible to represent the integral by$\sum_{k=1}^ma_k\lambda (A_k)$. Is this correct? How do I show it? Apparently I am missing something obvious since the author seems to deem it unnecessary to prove.

Answer 1

Usually notation is

$$\chi_A(x)= \left\{ \array{1 \quad \text{for } x \in A \\ 0 \quad \text{for }x \notin A }\right.$$

This is called the characteristic function of $A$.

To see it's well defined you need, e.g., to show that if you have another representation of $s$ then you get the same integral (if, e.g., $A = B\cup C$ you can write $\chi_{A}=\chi_B+\chi_C$. But of course more complicated situations can arise).

I deliberately do not show this since my understanding of your question is that you want to know what you have to show and want to try on your own.

Answer 2

First, it is $\chi$, where, $\chi_A(x)=1$ for $x\in A$ and $0$ elsewhere.

Second, the integral for simple function is well-defined because if we have simple function $s$ and $t$ but $s=t$, even they have different representation, we always can make them to have same representation, hence the integral is same

Answer 3

he is assuming first the simple functions. A simple function is a combination of characteristic functions, where $\chi_A(x)=1$ if $x\in A$ and $0$ elsewhere. Now clearly a characteristic function is measurable. So, the definition goes. Now observe that you can make the sets appearing in the definitions, disjoint. use this to show that the definition is well defined.