We prove the Dominated Convergence Theorem using Fatou's Lemma.

Given: $f_n \to f$ a.e., and $|f_n| \leq g$ where $g$ is integrable.

To prove: $\int_X f \, d\mu = \lim_{n \to \infty} \int_X f_n \, d\mu$

Proof:

Step 1: First, $|f| \leq g$ a.e., so $f$ is integrable. Also, $g \pm f_n \geq 0$ and $g \pm f \geq 0$.

Step 2: We have $(g - f_n) \to (g - f)$ a.e. Since $g - f_n \geq 0$, by Fatou's Lemma: $\int_X (g - f) \, d\mu \leq \liminf_{n \to \infty} \int_X (g - f_n) \, d\mu$

Step 3: Since $g$ is integrable: $\int_X g \, d\mu - \int_X f \, d\mu \leq \liminf_{n \to \infty} \left(\int_X g \, d\mu - \int_X f_n \, d\mu\right)$

$\int_X g \, d\mu - \int_X f \, d\mu \leq \int_X g \, d\mu - \limsup_{n \to \infty} \int_X f_n \, d\mu$

Subtracting $\int_X g \, d\mu$ (which is finite): $- \int_X f \, d\mu \leq - \limsup_{n \to \infty} \int_X f_n \, d\mu$

Thus: $\limsup_{n \to \infty} \int_X f_n \, d\mu \leq \int_X f \, d\mu$

Step 4: Similarly, applying Fatou's Lemma to $g + f_n \to g + f$: $\int_X (g + f) \, d\mu \leq \liminf_{n \to \infty} \int_X (g + f_n) \, d\mu$

$\int_X g \, d\mu + \int_X f \, d\mu \leq \int_X g \, d\mu + \liminf_{n \to \infty} \int_X f_n \, d\mu$

Thus: $\int_X f \, d\mu \leq \liminf_{n \to \infty} \int_X f_n \, d\mu$

Step 5: Combining Steps 3 and 4: $\int_X f \, d\mu \leq \liminf_{n \to \infty} \int_X f_n \, d\mu \leq \limsup_{n \to \infty} \int_X f_n \, d\mu \leq \int_X f \, d\mu$

Therefore all inequalities are equalities, and $\lim_{n \to \infty} \int_X f_n \, d\mu$ exists and equals $\int_X f \, d\mu$.