Lebesgue Integration - Applications

TheoremDominated Convergence Theorem (DCT)

Let $(X, \mathcal{F}, \mu)$ be a measure space. Suppose $\{f_n\}$ is a sequence of measurable functions such that:

$f_n(x) \to f(x)$ for almost every $x \in X$
There exists an integrable function $g: X \to [0, \infty)$ such that $|f_n(x)| \leq g(x)$ for all $n$ and almost every $x$

Then $f$ is integrable and: $\lim_{n \to \infty} \int_X f_n \, d\mu = \int_X f \, d\mu$

Moreover, $\lim_{n \to \infty} \int_X |f_n - f| \, d\mu = 0$ .

The Dominated Convergence Theorem is the most widely used convergence result in analysis. It requires pointwise convergence and a dominating function, but does not require monotonicity or uniform convergence.

ExampleDifferentiating Under the Integral

Let $f(x, t) = e^{-tx}$ for $x \in [0, \infty)$ and $t > 0$ . Define $F(t) = \int_0^\infty e^{-tx} \, dx = 1/t$ .

To show $F'(t) = \int_0^\infty \frac{\partial}{\partial t} e^{-tx} \, dx$ , consider: $\frac{F(t + h) - F(t)}{h} = \int_0^\infty \frac{e^{-x(t+h)} - e^{-xt}}{h} \, dx$

As $h \to 0$ , the integrand converges to $-xe^{-tx}$ . For $|h| < t/2$ , we have: $\left|\frac{e^{-x(t+h)} - e^{-xt}}{h}\right| \leq xe^{-xt/2}$

which is integrable. By DCT: $F'(t) = \int_0^\infty (-x) e^{-tx} \, dx = -\frac{1}{t^2}$

Remark

Applications of DCT include:

Leibniz integral rule: Under appropriate conditions, DCT justifies differentiating under the integral sign: $\frac{d}{dt} \int_X f(x, t) \, d\mu(x) = \int_X \frac{\partial f}{\partial t}(x, t) \, d\mu(x)$
Continuity of integrals: If $f_t(x) \to f_{t_0}(x)$ as $t \to t_0$ and $|f_t| \leq g$ integrable, then: $\lim_{t \to t_0} \int_X f_t \, d\mu = \int_X f_{t_0} \, d\mu$
$L^1$ convergence: The condition $\int_X |f_n - f| \, d\mu \to 0$ is equivalent to convergence in the $L^1$ norm, showing that pointwise a.e. convergence with domination implies $L^1$ convergence.

DCT is typically proved using Fatou's Lemma applied to $g + f_n$ , $g - f_n$ , and the fact that $g$ is integrable provides the necessary control to pass limits through the integral.