Definition of Lebesgue integral via improper Riemann integral.

Question

In the textbook Analysis by Lieb and Loss, the Lebesgue integral of a non-negative measurable function is defined as follows:

\begin{equation} \tag{1} \int_{\mathbb{R}^n} f(x) \: \mathrm{d}x := \int_{0}^{\infty} F_f(t) \: \mathrm{d}t, \end{equation} where $ \mathrm{d}x $ is the Lebesgue measure and $ F_f(t) := m(\{x \in \mathbb{R}^n | \: f(x) > t\}) $. So $ F_f(t) $ is a mapping from $ \mathbb{R}^+ $ to $ \mathbb{R}^+ \cup \{\infty\}$.

[N.B. I replaced the more general $ \Omega $ by $ \mathbb{R}^n $ and the general measure $ \mu(\mathrm{d}x) $ by the Lebesgue measure.]

Question 1: Now, my first question is whether (1) is to mean in detail that \begin{equation} \tag{2} \int_{\mathbb{R}^n} f(x) \: \mathrm{d}x := \lim_{a \to 0} \lim_{b \to \infty} \int_{a}^{b} F_f(t) \: \mathrm{d}t, \end{equation} since for a non-negative $ L^p $ function $ F_f(0) $ may well be equal to $ \infty $.

It also seems to me that we need both these limits (not only $ b \to \infty $, but also $ a \to 0 $) to make use of the fact that monotone functions (and $ F_f(0) $ is clearly one of them) are Riemann integrable on closed intervals.

Question 2: But then, how can we be sure that the double limit on the right side in (2) exists, i.e. that it takes a finite number or converges to $ \infty $. That is, how to be sure that the Riemann integral on the right side in (1) exists in the extended sense?

Any hints, suggestions or answers (as well as references to textbooks) would be much appreciated!

Answer 1

1. One can understand the right side as an improper Riemann integral if desired, in which case you would indeed use this formalism. But if you used the 1D Lebesgue integral to define the right side as well, then $F_f(0)=\infty$ is not a problem.

2. It's an integral of a nonnegative function, so it is a monotone function of $a,b$ (decreasing with $a$, increasing with $b$).

Answer 2

The Lebesgue integral of a function is defined over all of the space and is then limited to intervals by $$\int_a^b f := \int \chi_{(a,b)} f$$

Thus, $$\int_0^\infty F_f(t) \, dt = \int \chi_{(0,\infty)} F_f(t) \, dt$$