KindiakMath

A Less-Dominated Convergence Theorem

Problem 1. Let \{f_n\} be a sequence of functions defined on [0,\infty) satisfying the following properties:

  • for any interval of the form [0, b], f_n \to f uniformly on [a, b],
  • the improper integrals \displaystyle \int_0^\infty f_n and \displaystyle \int_0^\infty f converge,
  • there exists a nonnegative integrable function g such that |f_n| \leq g,
  • \displaystyle \int_0^\infty g converges.

Prove that \displaystyle \lim_{n \to \infty}\int_0^\infty f_n = \int_0^\infty f.

Remark 1. This is a weak form of the dominated convergence theorem. The strong form which requires measure-theoretic machinery only requires point-wise convergence.

(Click for Solution)

Solution. Fix \epsilon > 0. Our line of attack is to find sufficiently large M > 0 so that integrals of the form \displaystyle \int_M^\infty can be controlled due to the integrals converging, while integrals of the form \displaystyle \int_0^M can be controlled using the uniform convergence of f_n \to f.

To that end, let’s first ascertain that |f| \leq g point-wise. Fix x \geq 0. Applying the uniform convergence f_n \to f on [0, x], for any k_1 > 0, there exists N_1 \in \mathbb N such that

n > N_1 \quad \Rightarrow \quad \|f_n - f\|_\infty < k_1 \cdot \epsilon.

This implies the upper-found

|f(x)| \leq |f_n(x)| + |f_n(x) - f(x)| < g(x) +k_1 \cdot \epsilon.

Since \epsilon > 0 is arbitrary, |f| \leq g point-wise. Therefore,

|f_n - f| \leq |f_n|+|f| \leq 2g

point-wise. The motivation for this step is to do some upper bounds on tail integrals. Since \displaystyle \int_0^\infty g converges, for any k_2 > 0, there exists M > 0 such that

\displaystyle \int_M^\infty g < k_2 \cdot \epsilon.

Employing our estimate from previous discussions,

\displaystyle \int_M^\infty |f_n - f| \leq 2 \int_M^\infty g = (2 k_2) \cdot \epsilon.

Applying the uniform convergence f_n \to f on [0, M], for any k_3 > 0, there exists N_2 \in \mathbb N such that

n > N_2 \quad \Rightarrow \quad \|f_n - f\|_\infty < k_3 \cdot \epsilon.

This means for the infinite integrals, whenever n > N_2,

\displaystyle \begin{aligned} \left| \int_0^\infty f_n - \int_0^\infty f \right| &\leq \int_0^\infty |f_n - f|\\ &\leq \int_0^M |f_n - f| + \int_M^\infty |f_n - f| \\ &\leq \int_0^M (k_3 \cdot \epsilon) + (2 k_2) \cdot \epsilon \\ &= (M \cdot k_3 + 2 \cdot k_2) \cdot \epsilon.\end{aligned}

Taking care of the dependencies, setting k_2 = 1/4, k_3 = 1/(4M) yields the desired final result:

\displaystyle \lim_{n \to \infty}\int_0^\infty f_n = \int_0^\infty f.

—Joel Kindiak, 31 Jan 25, 1213H

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