Simple Functions

Setting: Measure space $(\Omega, \mathcal{F}, \mu)$. Note that $\mu$ is not necessarily a probability measure.

We assume $\mu$ is $\sigma$-finite, meaning that: $\exists A_1, A_2, \ldots \in \mathcal{F}$ such that $\mu(A_n) < \infty \ \forall n$ and $\bigcup_{n=1}^\infty A_n = \Omega$.

Goal: Define the Lebesgue integral for general functions, building from simple functions.

Definition: Simple Function

A function $\varphi$ is called a simple function if:

$$\varphi(\omega) = \sum_{i=1}^n a_i \mathbb{1}_{A_i}$$

where $a_i \in \mathbb{R}$, $A_i \in \mathcal{F}$, and $\mu(A_i) < \infty$ for all $i$. Here $\mathbb{1}_{A_i}$ is the indicator function of the set $A_i$.

Simple functions are the Lebesgue counterpart of step functions used for Riemann integrals.

In the Lebesgue setting, we don’t need the domain to be the real line or partitioned into intervals. Any measurable subset $A_i$ can serve as the base for a constant value.

In this simple case, the integration is just the volume (height $\times$ size of set) of the region under the function, i.e.,

$$\int \varphi \, d\mu = a_1 \mu(A_1) + a_2 \mu(A_2)$$

Definition: Lebesgue Integral of a Simple Function

For a simple function $\varphi = \sum_{i=1}^n a_i \mathbb{1}_{A_i}$, the integral is defined as:

$$\int \varphi \, d\mu = \sum_{i=1}^n a_i \mu(A_i)$$

(Note: Usually we assume $A_i$ are disjoint and partition $\Omega$, i.e., $\bigcup_i A_i = \Omega$).

Properties

Lemma: Properties of Simple Integral

Statement

Let $\varphi, \psi$ be two simple functions. Then:

1. $\varphi \ge 0$ a.e. $\implies \int \varphi \, d\mu \ge 0$.

   Note: “a.e.” (almost everywhere) means the property holds except on a set of measure zero: $\mu( \{ \omega : \varphi(\omega) < 0 \} ) = 0$. In probability settings, this corresponds to a.s. (almost surely).

2. $\int a\varphi \, d\mu = a \int \varphi \, d\mu$ for any $a \in \mathbb{R}$.
3. $\int (\varphi + \psi) \, d\mu = \int \varphi \, d\mu + \int \psi \, d\mu$.
4. $\varphi \le \psi$ a.e. $\implies \int \varphi \, d\mu \le \int \psi \, d\mu$.
5. $\varphi = \psi$ a.e. $\implies \int \varphi \, d\mu = \int \psi \, d\mu$.
6. $|\int \varphi \, d\mu| \le \int |\varphi| \, d\mu$.

Proof

1. Assume $\varphi = \sum_{i=1}^n a_i \mathbb{1}_{A_i}$ where $A_i$ are disjoint. If $\varphi \ge 0$ a.e., then $a_i \ge 0$ for all $i$ where $\mu(A_i) > 0$. Since $\int \varphi \, d\mu = \sum a_i \mu(A_i)$ and each term is $a_i \mu(A_i) \ge 0$, the sum is $\ge 0$.

2. For $\varphi = \sum a_i \mathbb{1}_{A_i}$, we have $a\varphi = \sum (a a_i) \mathbb{1}_{A_i}$. By definition, $\int a\varphi \, d\mu = \sum (a a_i) \mu(A_i) = a \sum a_i \mu(A_i) = a \int \varphi \, d\mu$.

3. Suppose $\varphi = \sum_i a_i \mathbb{1}_{ A_i }$ and $\psi = \sum_j b_j \mathbb{1}_{ B_j }$, where the $A_i$ and $B_j$ partition $\Omega$ respectively.

   Consider the intersection $A_i \cap B_j$. On this set, $\varphi + \psi = a_i + b_j$. Note that $\bigcup_i \bigcup_j ( A_i \cap B_j ) = \Omega$ and these sets are disjoint. Thus:

   $$\begin{aligned} \int ( \varphi + \psi ) \, d\mu &= \sum_i \sum_j ( a_i + b_j ) \mu( A_i \cap B_j ) \\ &= \sum_i \sum_j a_i \mu( A_i \cap B_j ) + \sum_i \sum_j b_j \mu( A_i \cap B_j ) \\ &= \sum_i a_i \left( \sum_j \mu( A_i \cap B_j ) \right) + \sum_j b_j \left( \sum_i \mu( A_i \cap B_j ) \right) \\ &= \sum_i a_i \mu( A_i ) + \sum_j b_j \mu( B_j ) \\ &= \int \varphi \, d\mu + \int \psi \, d\mu \end{aligned}$$

4. Since $\psi - \varphi \ge 0$ a.e., by property 1 and 3: $\int \psi \, d\mu - \int \varphi \, d\mu = \int (\psi - \varphi) \, d\mu \ge 0$.

5. Follows directly from 4 by applying both $\varphi \le \psi$ and $\psi \le \varphi$.

6. Observe that $|\varphi| = \max(\varphi, -\varphi)$. Since $\varphi \le |\varphi|$, we have $\int \varphi \, d\mu \le \int |\varphi| \, d\mu$ by (4). Since $-\varphi \le |\varphi|$, we have $-\int \varphi \, d\mu = \int (-\varphi) \, d\mu \le \int |\varphi| \, d\mu$. Together, these imply $|\int \varphi \, d\mu| \le \int |\varphi| \, d\mu$.