Convergence

A central question in integration theory is dealing with limits: Limit of integration vs. Integration of limit. Under what conditions can we interchange them?

$$\lim_{n \to \infty} \int f_n \, d\mu \stackrel{?}{=} \int \lim_{n \to \infty} f_n \, d\mu$$

Monotone Convergence Theorem (MCT)

Theorem: Monotone Convergence

Statement

If $f_n \ge 0$ and $f_n \uparrow f$ (non-decreasing sequence converging to $f$), then:

$$\int f_n \, d\mu \uparrow \int f \, d\mu$$

Proof Idea

1. $\le$ direction: Since $f_n \le f_{n+1} \le f$ for all $n$, by monotonicity of the integral:

   $$\int f_n \, d\mu \le \int f \, d\mu$$

   Thus $\lim \int f_n \, d\mu \le \int f \, d\mu$.

2. $\ge$ direction: Let $\varphi$ be any simple function such that $0 \le \varphi \le f$. Let $c \in (0, 1)$. Consider the sets $E_n = \{ x : f_n(x) \ge c \varphi(x) \}$. Since $f_n \uparrow f$ and $f \ge \varphi$, the sets $E_n$ grow to cover the whole space where $\varphi > 0$.

   $$\int f_n \, d\mu \ge \int_{E_n} f_n \, d\mu \ge c \int_{E_n} \varphi \, d\mu$$

   Taking limit $n \to \infty$: $\lim \int f_n \ge c \int \varphi$. Letting $c \to 1$: $\lim \int f_n \ge \int \varphi$. Since this holds for any simple $\varphi \le f$, taking the supremum gives $\lim \int f_n \ge \int f$.

Fatou’s Lemma

Fatou's Lemma

Statement

If $f_n \ge 0$ for $n = 1, 2, \dots$, then:

$$\liminf_{n \to \infty} \int f_n \, d\mu \ge \int \left( \liminf_{n \to \infty} f_n \right) \, d\mu$$(intuitive)

Proof Idea

1. Define $g_k(x) = \inf_{n \ge k} f_n(x)$. Then $g_k$ is a non-decreasing sequence: $g_1 \le g_2 \le \dots$

2. Also, by definition of limit inferior: $\lim_{k \to \infty} g_k = \sup_k (\inf_{n \ge k} f_n) = \liminf_{n \to \infty} f_n$.

3. Since $g_k \le f_n$ for all $n \ge k$, we have $\int g_k \, d\mu \le \int f_n \, d\mu$ for all $n \ge k$.

   Therefore, $\int g_k \, d\mu \le \inf_{n \ge k} \int f_n \, d\mu$.

4. Apply the Monotone Convergence Theorem to $g_k$:

   $$\begin{aligned} \int \liminf f_n \, d\mu &= \int \lim g_k \, d\mu = \lim \int g_k \, d\mu \\ &\le \lim_{k \to \infty} \left( \inf_{n \ge k} \int f_n \, d\mu \right) \\ &= \liminf_{n \to \infty} \int f_n \, d\mu \end{aligned}$$

Dominated Convergence Theorem (DCT)

Theorem: Dominated Convergence

Statement

Suppose $f_n \to f$ a.e. (almost everywhere). If there exists an integrable function $g$ such that:

$$|f_n| \le g \quad \text{for all } n$$

(we say $f_n$ is dominated by $g$), then:

$$\int f_n \, d\mu \to \int f \, d\mu$$

Proof Idea

Since $|f_n| \le g$, we have $-g \le f_n \le g$, thus $g + f_n \ge 0$ and $g - f_n \ge 0$. Apply Fatou’s Lemma to these non-negative sequences.

1. For $g + f_n$:

   $$\begin{aligned} \int (g+f) &= \int \liminf (g+f_n) \\ &\le \liminf \int (g+f_n) \\ &= \int g + \liminf \int f_n \end{aligned}$$

   Subtracting $\int g$ (which is finite) gives: $\int f \le \liminf \int f_n$.

2. For $g - f_n$:

   $$\begin{aligned} \int (g-f) &= \int \liminf (g-f_n) \\ &\le \liminf \int (g-f_n) \\ &= \int g + \liminf \int (-f_n) \\ &= \int g - \limsup \int f_n \end{aligned}$$

   Subtracting $\int g$ gives: $-\int f \le -\limsup \int f_n \implies \int f \ge \limsup \int f_n$.

3. Conclusion:

   $$\limsup \int f_n \le \int f \le \liminf \int f_n$$

   This implies the limit exists and equals $\int f$.

Summary

So, if the sequence of functions $f_n$ is monotone or dominated, then we can “push” the limit of integration inside the integral.