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Measure-Theoretic Integration#

Lebesgue Integrals#

Consider a subset \(S \subseteq \mathbb{R}\) and its indicator function \(\mathbf{1}_S: \mathbb{R} \to \mathbb{R}\). The graph of \(\mathbf{1}_S\) would resemble the following, since \(\mathbf{1}_S(x) = 1\) if \(x \in S\) and \(\mathbf{1}_S(x) = 0\) otherwise:

The area under the graph of \(\mathbf{1}_S\) is just the sum of the shaded areas, all of which have a height of \(1\). The area of each of these is given by \(1 \cdot \mu (S_i)\), where \(\mu(S_i)\) is the size (length) of each blue piece. Therefore, the area under the graph of \(\mathbf{1}_S\) is \(\sum_{i} 1 \cdot \mu(S_i) = 1\cdot \sum_{i} \mu(S_i) = \sum_{i} \mu(S_i)\). However, the sum \(\sum_{i} \mu(S_i)\) of the sizes of the blue pieces \(S_i\) is just equal to the size \(\mu(S)\) of \(S\). Thus we define the integral of \(\mathbf{1}_S\) in the following way.

Definition: Lebesgue Integral of Indicator Functions

Let \((X, \Sigma, \mu)\) be a measure space, let \(S \subseteq X\) be measurable and let \(\mathbf{1}_S: X \to \mathbb{R}\) be the indicator function of \(S\).

The (Lebesgue) integral of \(\mathbf{1}_S\) with respect to \(\mu\) is the measure of \(S\):

\[ \int \mathbf{1}_S \mathop{\mathrm{d}\mu} \overset{\text{def}}{=} \mu(S) \]

We now extend this definition to functions which can be built from finitely many indicator functions.

Definition: Simple Function

Let \((X, \Sigma, \mu)\) be a measure space and let \(\phi: \mathcal{D} \subseteq X \to \mathbb{R}_{\ge 0}\) be a real-valued function.

We say that \(\phi\) is a simple function if \(\mathcal{D}\) can be represented as the union of a finite collection disjoint measurable sets \(\mathcal{D} = \mathcal{D}_1 \cup \cdots \cup \mathcal{D}_n\) and there exist \(n\) real numbers \(c_1, \dotsc, c_n \ge 0\) such that

\[ \phi(x) = \sum_{i = 1}^n c_i \mathbf{1}_{\mathcal{D}_i}(x) \qquad \forall x \in \mathcal{D}, \]

where \(\mathbf{1}_{\mathcal{D}_i}\) is the indicator function of \(\mathcal{D}_i\).

Consider a simple real function \(\phi: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\).

The are under its graph is just the sum of the shaded areas. All shaded areas with a blue bottom have a height of \(c_1\) and all shaded areas with a green bottom have a height of \(c_2\). Applying the same reasoning as we did with indicator functions, the total shaded area with a blue bottom is \(c_1 \cdot \mu (\mathcal{D}_1)\). Similarly, the total shaded area with a green bottom is \(c_2 \cdot \mu(\mathcal{D}_2)\). The area under the graph of \(\phi\) is then given by summing these two areas: \(c_1 \cdot \mu (\mathcal{D}_1) + c_2 \cdot \mu(\mathcal{D}_2)\). More generally, when \(\mathcal{D}\) must be decomposed into \(n\) subsets \(\mathcal{D} = \mathcal{D}_1 \cup \cdots \cup \mathcal{D}_n\), the total area is given by \(\sum_{i=1}^n c_i \cdot \mu(\mathcal{D}_i)\).

Definition: Lebesgue Integral of Simple Functions

Let \((X, \Sigma, \mu)\) be a measure space and let \(\phi: \mathcal{D} \subseteq X \to \mathbb{R}_{\ge 0}\) be a real-valued function such that \(\mathcal{D}\) can be represented as the union of a finite collection disjoint measurable sets \(\mathcal{D} = \mathcal{D}_1 \cup \cdots \cup \mathcal{D}_n\) and there exist \(n\) real numbers \(c_1, \dotsc, c_n \ge 0\) with

\[ \phi(x) = \sum_{i = 1}^n c_i \mathbf{1}_{\mathcal{D}_i}(x) \qquad \forall x \in \mathcal{D}. \]

The (Lebesgue) integral of \(\phi\) with respect to \(\mu\) is defined using the measures of \(\mathcal{D}_i\) as

\[ \int \phi \mathop{\mathrm{d}\mu} \overset{\text{def}}{=} \sum_{i = 1}^n c_i \mu(\mathcal{D}_i), \]

with the convention \(0 \cdot \infty = 0\).

This definition can be extended to non-negative real-valued functions. Consider a real function \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}_{\ge 0}\). We can approximate \(f\) using a simple function. Choose \(n \in \mathbb{N}\) values \(c_1, \dotsc, c_n \in f(\mathcal{D})\) from the range of \(f\) and let \(\mathcal{D}_i = \{x \in \mathcal{D} \mid f(x) \ge c_i\}\). Define \(\phi: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}_{\ge 0}\) in the following way:

\[ \phi(x) \overset{\text{def}}{=} \begin{cases}c_1 \qquad \text{if } x \in \mathcal{D}_1 \\ \vdots \\ c_n \qquad \text{if } x \in \mathcal{D}_n\end{cases} \]

We see \(\phi\) is a simple function

\[ \phi(x) = \sum_{i = 1}^n c_i \mathbf{1}_{\mathcal{D}_i}(x), \]

since \(\mathbf{1}_{\mathcal{D}_j}(x)\) gives \(1\) only when \(x \in \mathcal{D}_j\) and gives \(0\) otherwise. Therefore, \(\phi(x) = c_j\) if and only if \(x \in \mathcal{D}_j\). Of course, if we choose only a few values \(c_1, \cdots, c_n\), then \(\phi\) will be a pretty bad approximation of \(f\). However, as we increase the number of values \(n\), \(\phi\) becomes a better and better approximation. This is illustrated in the following animation:

Lines with the same color correspond to the same value \(c_j\) and \(\mathcal{D}_j\) is the part of the of \(\mathbb{R}\) which lies directly below the lines corresponding to \(c_j\).

The more closely \(\phi\) approximates \(f\), the more the area under the graph of \(\phi\) resembles the area under the graph of \(f\):

This notion of \(\phi\) approximating \(f\) can be defined and proven rigorously via function sequences and their limits but that is unnecessary. Notice that \(0 \le \phi(x) \le f(x)\) for all \(x \in \mathcal{D}\). Consider now the set \(S\) of the integrals of all simple functions \(s: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) such that \(0 \le s(x) \le f(x)\) for all \(x \in \mathcal{D}\). Since these simple functions are always less than or equal to \(f\), the areas under their graphs must be less than or equal to the area under the graph of \(f\). The supremum of \(S\), i.e. the largest of these areas, is thus as close as one could get to the area under the graph of \(f\). Moreover, it can be proven that this is equal to the integral of \(\phi\) as \(\phi\) becomes a better approximation of \(f\).

Definition: Integration of Non-Negative Functions

Let \((X, \Sigma, \mu)\) be a measure space and let \(f: \mathcal{D} \to \mathbb{R}_{\ge 0}\) be a measurable (in the sense of the Lebesgue measure) real-valued function on a measurable subset \(\mathcal{D} \subseteq X\).

The integral of \(f\) is the supremum of the set of all integrals of simple functions \(s\) such that \(0 \le s \le f\):

\[ \int f \mathop{\mathrm{d}\mu} \overset{\text{def}}{=} \sup \left\{ \int s \mathop{\mathrm{d}s} \mid s \text{ is simple and } 0 \le s \le f\right\} \]

Extending the above definition to real-valued functions which can also take on negative values is fairly easy. Consider a real function \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\). Essentially, we split \(f\) into a negative and a non-negative part. We take the area between the horizontal axis and the graph of \(f\)'s non-negative part of \(f\) and subtract from it the area between the horizontal axis and the graph of \(f\)'s negative part. This gives us a notion of the signed area between the horizontal axis and the graph of \(f\) and thus yields a good definition for the integral of \(f\).

Definition: Integration of General Functions

Let \((X, \Sigma, \mu)\) be a measure space, let \(f: \mathcal{D} \to \mathbb{R}\) be a measurable (in the sense of the Lebesgue measure) real-valued function on a measurable subset \(\mathcal{D} \subseteq X\) and let \(f^{-} = \max (-f, 0)\) and \(f^{+} = \max (f, 0)\).

The (Lebesgue) integral of \(f\) with respect to \(\mu\) is defined using the integrals of \(f^{-}\) and \(f^{+}\) as

\[ \int f \mathop{\mathrm{d}\mu} \overset{\text{def}}{=} \int f^{+} \mathop{\mathrm{d}\mu} - \int f^{-} \mathop{\mathrm{d}\mu}, \]

provided that at least one of \(\int f^{+} \mathop{\mathrm{d}\mu}\) or \(\int f^{-} \mathop{\mathrm{d}\mu}\) is finite (in order to avoid \(\infty - \infty\)).

We say that \(f\) is \(\mu\)-integrable if its integral is finite.

Definition: Integrating on a Subset

Let \(S\) be a measurable subset \(S \subseteq X\).

The (Lebesgue) integral of \(f\) over \(S\) with respect to \(\mu\) is the integral

\[ \int_{S} f \mathop{\mathrm{d}\mu} \overset{\text{def}}{=} \int f \cdot \mathbf{1}_S \mathop{\mathrm{d}\mu}, \]

where \(\mathbf{1}_S\) is the indicator function of \(S\).

We say that \(f\) is \(\mu\)-integrable on \(S\) if its integral on \(S\) is finite.

Note: Lebesgue Integral with Respect to the Lebesgue Measure

When \(X\) is \(\mathbb{R}^n\) (or \(\mathbb{R}\)) and \(\mu\) is the Lebesgue measure on \(\mathbb{R}^n\) (or \(\mathbb{R}\)), then we just say "Lebesgue integral" and omit "with respect to". Similarly, we just say that \(f\) is "Lebesgue-integrable".

Theorem: Integrability Criterion

Let \((X, \Sigma, \mu)\) be a measure space, let \(f: \mathcal{D} \to \mathbb{R}\) be a measurable (in the sense of the Lebesgue measure) real-valued function on a measurable subset \(\mathcal{D} \subseteq X\) and let \(S\) be a measurable subset \(S \subseteq X\).

The function \(f\) is \(\mu\)-integrable on \(S\) if and only if the integral of its absolute value is \(\mu\)-integrable on \(S\):

\[ \int_S f \mathop{\mathrm{d}\mu} \lt \infty \iff \int_S |f| \mathop{\mathrm{d}\mu} \lt \infty \]
Proof

TODO