Non-Negative Functions

Now we don’t need the functions to be bounded or supported on a set with finite measure. We extend the definition of the integral to non-negative measurable functions ($f \ge 0$).

Definition

Let $f \ge 0$ be a measurable function. We define:

$$\int f \, d\mu = \sup \left\{ \int h \, d\mu : 0 \le h \le f, \ h \in H \right\}$$

where $H$ denotes the class of functions that are bounded and supported on a set with finite measure.

Approximation Strategy

One way is to approximate $f$ by a sequence of truncated functions. Define:

$$h_n = (f \wedge n) \mathbf{1}_{E_n}$$

where $E_n \uparrow \Omega$ is a sequence of sets with $\mu(E_n) < \infty$ for all $n$; and $f \wedge n = \min(f, n)$.

We approximate $f$ by clipping its height at $n$ and restricting its domain to a set $E_n$ of finite measure. Shown: $h_1 = (f \wedge 1) \mathbf{1}_{E_1}$ and $h_2 = (f \wedge 2) \mathbf{1}_{E_2}$.

Lemma

Statement

$$\lim_{n \to \infty} \int h_n \, d\mu = \int f \, d\mu$$

Proof

1. Existence of Limit: Since $n$ increases and $E_n$ expands, $h_n$ is a non-decreasing sequence ($h_n \le h_{n+1}$). Thus $\int h_n \, d\mu$ is non-decreasing, so the limit $\lim_{n \to \infty} \int h_n \, d\mu$ exists.

2. Lower Bound: Let $h \in H$ be any function in the class defined above. Let $M$ be its bound ($h \le M$). Since $h$ is supported on a set of finite measure, let $E = \{x : h(x) > 0\}$ with $\mu(E) < \infty$.

   For any $n \ge M$:

   $$\int_{E_n} h_n \, d\mu = \int_{E_n} (f \wedge n) \, d\mu \ge \int_{E_n} h \, d\mu$$

   (since on $E_n$, $h_n = f \wedge n \ge h \wedge n = h$ because $h \le f$ and $h \le M \le n$).

3. Decomposition:

   $$\int_{E_n} h \, d\mu = \int h \, d\mu - \int_{E_n^c} h \, d\mu$$

   Consider the term $\int_{E_n^c} h \, d\mu$. Since $h$ is supported on $E$:

   $$\int_{E_n^c} h \, d\mu = \int_{E_n^c \cap E} h \, d\mu \le \int_{E_n^c \cap E} M \, d\mu = M \mu(E \cap E_n^c) = M \mu(E \setminus E_n)$$

   As $E_n \uparrow \Omega$, we have $E \setminus E_n \downarrow \emptyset$, so $\mu(E \setminus E_n) \to 0$ (since $\mu(E) < \infty$). Thus, $\int_{E_n^c} h \, d\mu \to 0$.

4. Inequality: So we have $\lim_{n \to \infty} \int h_n \, d\mu \ge \int h \, d\mu$. Since this holds for any $h \in H$, taking the supremum over all such $h$:

   $$\lim_{n \to \infty} \int h_n \, d\mu \ge \sup_{h \in H} \int h \, d\mu = \int f \, d\mu$$

5. Equality: On the other hand, since each $h_n \in H$ (it is bounded by $n$ and supported on $E_n$), by definition:

   $$\int h_n \, d\mu \le \int f \, d\mu \implies \lim_{n \to \infty} \int h_n \, d\mu \le \int f \, d\mu$$

   Combining both inequalities, we get the result.

Important

This proof does not imply that the integral is finite. Both $\lim \int h_n$ and $\int f$ can be $+\infty$.

Properties

Exercise 7

Problem

Prove that Properties 1-6 also hold for non-negative measurable functions.

Hint

Use the properties for bounded functions and apply the approximation strategy.