Well defined function on Lebesgue integration

202 Views Asked by At

I have a partial definition for nonnegative functions. $S$ denotes the class of simple functions $s$ on $\mathbb R^n$ such that $0\leq s(x)<\infty$.

If $s\in S$, then the integral of $s$, denoted $\int sd\lambda$, is the number given as follows: if $s$ is presented in the form $$s=\sum_{k=1}^m \alpha_k\chi_{A_k}$$ where $0\leq \alpha_k<\infty$ and the sets $A_k$ are measurable and disjoint, then

$$\int sd\lambda =\sum_{k=1}^m \alpha_k \lambda (A_k)$$

Assuming I know that $\int sd\lambda$ is well defined, how can I show that it follows

  • $0\leq\int sd\lambda \leq \infty$?
  • If $0\leq c<\infty$ is a constant, $\int csd\lambda = c\int sd\lambda$?

Is it that obvious (written in a book)? Am I missing something?

1

There are 1 best solutions below

3
On BEST ANSWER

This is fairly obvious, if the integral is well defined.

Assuming $\lambda $ takes on only positive values or $\infty$, that is. In this case the sum which defines the integral is a finite sum of nonnegative terms, implying the first item.

Then, if $$s = \sum \alpha_k\chi_{A_k}$$ you have $$cs = \sum c\alpha_k\chi_{A_k}$$ so $$\int cs d\lambda = \sum c\alpha_k\lambda(A_k) =c \sum \alpha_k\lambda(A_k)=c\int s d\lambda$$