Intro to Lebesgue Integration

Abstract Integration Characteristic function The Integral of Finite Nonnegative Measurable Functions The Integral of Nonnegative Measurable Functions The Integral of General Measurable Functions

Abstract Integration

Lebesgue integration is sometimes referred to as abstract integration. The reason is that the Lebesgue integral is developed in the very general setting of a measure space \((\Omega, \mathcal{F}, \mu)\).

We want to define the integral: \[ \int_{\Omega} g(\omega) d\mu(\omega) \] of a measurable function \(g: \Omega \to \overline{\mathbb{R}}\) defined on a measure space \((\Omega, \mathcal{F}, \mu)\).
Note: \(\overline{\mathbb{R}}\) refers to the extended set of real values, which includes \(\infty\) and \(-\infty\).

Let's check some special cases:

If the measure space is a probability space:\((\Omega, \mathcal{F}, \mathbb{P})\), and \(X: \Omega \to \bar{\mathbb{R}}\) is measurable, which means \(X\) is an extended-valued random variable, then the integral is the expectation of \(X\): \[ \int_{\Omega} X \, d\mathbb{P} = \mathbb{E }(X). \] If we define the measure space as \((\mathbb{R}, \mathcal{B}, \lambda)\) where \(\mathcal{B}\) is the Borel \(\sigma\)-algebra and \(\lambda\) is he Lebesgue measure, then the integral is a generalization of the usual integral encountered in calculus: \[ \int_{\mathbb{R}} g \, d\lambda = \int g(x) dx. \]
So, even if it isn't always stated explicitly, many of the integrals we have encountered — whether in the context of computing expectations in probability theory or evaluating integrals on the real line — can be defined using Lebesgue integration. This approach is the core of modern analysis and probability, providing a robust framework for handling functions that may be too irregular or complex for the Riemann integration.

Characteristic function

The integral over a measurable subset \(B\) of \(g\) is defined by \[ \int_B g \, d\mu = \int (\chi_B \, g)d\mu \] where \(\chi_B\)(or alternatively, \( 1_B\)) is a characteristic function(or indicator function) so that \[ (\chi_B \, g) (\omega) = \begin{cases} g(\omega) & \text{if \(\omega \in B\)}\\ 0 & \text{if \(\omega \notin B\)}. \end{cases} \]
We will use the term "almost everywhere"(a.e.) to mean "for all \(\omega\) outside a zero-measure subset of \(\Omega\). So, we "ignore" a set of \(\omega\)'s that has measure zero. For the special case of probability measure, we also use "almost surely"(a.s.).

For example, \[ g_n \uparrow g, \, a.e. \] means that the increasing monotonic convergence of \(g_n(\omega)\) to \(g(\omega)\) holds \(\forall \omega \in \Omega\) outside a zero-measure set.

The Integral of Finite Nonnegative Measurable Functions

A function \(g: \Omega \to \mathbb{R}\) is said to be simple if it is measurable, finite and takes only finitely many different values.

If \(g\) is a simple function of the form: \[ g(\omega) = \sum_{i=1}^k a_i \chi_{A_i}(\omega), \quad \forall \omega \in \Omega \] where \(k\) is a finite nonnegative integer, \(a_i \in \mathbb{R}\) and \(A_i\) are measurable sets.
Then its integral is defined by \[ \int g \, d\mu = \sum_{i=1}^k a_i \mu(A_i). \] (We assume \(a_i \mu(A_i) = 0\) if \(a_i = 0\) and \(\mu(A_i) = \infty\).)

Example: Dirichlet function Finally, we can get the integral of Dirichlet function on the interval \([0, 1]\): \[ f(x)= \begin{cases} 1 &\text{if \(x \in \mathbb{Q}\)} \\ 0 &\text{if \(x \in \mathbb{R} \setminus \mathbb{Q}\)} \end{cases} \]
The Dirichlet function is a simple function since it only takes on the two values 0 and 1. By the definiton: \[ \begin{align*} \int_0^1 f(x)dx &= \int_0^1 \chi_{\mathbb{Q}}(x) dx\\\\ &= 1 \cdot \mu([0, 1] \cap \mathbb{Q})(x) + 0 \cdot \mu([0, 1] \setminus \mathbb{Q})(x)\\\\ &= 1 \cdot 0 + 0 \cdot (1 - 0) \\\\ &= 0 \end{align*} \] Note: Technically, for any interval \([a, b]\), the Lebesgue integral of the Dirichlet function is 0 because the Lebesgue measure of any countable set is zero.

The Integral of Nonnegative Measurable Functions

We approximate the integral of a nonnegative function \(g\) using simple functions.

For a nonnegative (extended-valued) measurable function \(g: \Omega \to [0, \infty]\), we let \(S(g)\) be the set of all nonnegative simple (hence automatically measurable) functions \(q\) that satisfy \(0 \leq q \leq g\), and define \[ \int g \, d\mu = \sup_{q \in S(g)} \int q \, d\mu. \]

The Integral of General Measurable Functions

Consider a measurable function \(g: \Omega \to \overline{\mathbb{R}}\). Let \[ A_+ = \{\omega | g(\omega) > 0\}, \qquad g_+ = g \cdot \chi_{A_+} \] and \[ A_- = \{\omega | g(\omega) < 0\}, \qquad g_- = g \cdot \chi_{A_-} \] Note that \(A_+\) , \(A_-\) are measurable sets, and \(g_+\) , \(g_-\) are nonnegative(possibly extended-valued) measurable functions.

Then we have \(g = g_+ - g_-\) and define \[ \int g \, d\mu = \int g_+ \, d\mu - \int g_- \, d\mu \] if we have both \(\int g_+ \, d\mu < \infty\) and \(\int g_- \, d\mu < \infty\).

Note: The definition implies there exists a function that is NOT Lebesgue integrable over some interval. In general, if a function is Riemann integrable, the function is also Lebesgue integrable and \[ \int_{[a, b]} f(x) dx = \int_a^b f(x) dx \] but some improperly Riemann integrable function cannot be Lebesgue integrable.

Example: Sinc function over \([0, \infty)\) Consider the Dirichlet integral: \[ \int_0^{\infty} \frac{\sin x}{x} dx \] This is known to converge to \(\frac{\pi}{2}\): \[ \begin{align*} \int_0^{\infty} \frac{\sin x}{x} dx &= \lim_{b \to \infty} \int_0^b \frac{\sin x}{x} dx \\\\ &= \frac{\pi}{2} \end{align*} \] So, \(f\) is improperly Riemann integrable on \([0, \infty)\).

On the other hand, in the Lebesgue sense: \[ \int_0^{\infty} \frac{\sin x}{x} dx = \int_0^{\infty} \left(\frac{\sin x}{x}\right)_+ dx - \int_0^{\infty} \left(\frac{\sin x}{x}\right)_- dx \] \(\sin x\) is always positive on the interval \([2\pi n, 2\pi n + \pi]\), and then \[ \begin{align*} \int_0^{\infty} \left(\frac{\sin x}{x}\right)_+ dx &= \sum_{n=0}^{\infty} \int_{2\pi n}^{2\pi n + \pi} \frac{\sin x}{x} dx \\\\ &\geq \sum_{n=0}^{\infty} \int_{2\pi n}^{2\pi n + \pi} \frac{\sin x}{2 \pi n + \pi} dx \\\\ &= \sum_{n=0}^{\infty} \frac{2}{\pi(2n +1)} \\\\ &= \infty \end{align*} \] and similarly, \(\int_0^{\infty} \left(\frac{\sin x}{x}\right)_- dx = \infty\). Thus, by the definition, this is NOT integrable in the Lebesgue sense.