Lebesgue integral of a non-negative function.

Question

I have been looking at Kolmogorov's book "Introductory Real Analysis" and have become stuck at the problem 4a) on page 301.

In this problem we are given $f$ a nonnegative and integrable function on $A$, a set of finite measure. We are asked to show that

$\int_A f(x) d \mu \ge 0$

However, how would we prove this? In this text the Lebesgue integral of a function $f$ is defined to be (when the limit exists);

$\int_A f(x) d\mu = \lim\limits_{n \rightarrow \infty} \int_{A} f_n(x) d\mu$

where $f_n$ is a sequence of integrable simple functions that converges uniformly to $f$.

I have attempted to show this by showing that the sequence of simple functions converging to a nonnegative function, is a sequence of nonnegative functions. However this isn't true, take $f_n = -1/n$ and $f = 0$. How else could I go about this?

Thanks.

Answer 1

Hint: First show the result for non-negative simple functions. Then use that you can find $f_n \ge 0$ simple with $f_n \nearrow f$ uniformly to see that $$\int_A f d\mu = \lim_{n\to\infty} \underbrace{\int_A f_n d\mu}_{\ge 0} \ge 0$$ For simple functions, the result is fairly easy to prove but depends slightly on the concrete definition of a "simple function" in your book.

Answer 2

Simply state that you are restricting to a sequence simple functions that are non negative.

Or use the fact that $f\ge g\Rightarrow \int f\ge \int g $

Then $\int f = \sup\{\int s|s, simple\}=\sup\{\int s|s, simple,positive\}$