Here's an adaptation of Exercises 2, 3, and 4 from Section 2.4, Limit Theorems, in Donald Cohn's Measure Theory.

First, his version of the monotone convergence theorem and dominated convergence theorems.

Theorem (Monotone Convergence Theorem). Suppose $(X, \mathscr{A}, \mu)$ is a measure space, with $f, f_n: X \to [0, +\infty]$, $\mathscr{A}$-measurable functions. Suppose $f_n \uparrow f$ a.e. Then $\int f_n \to \int f$.

Theorem (Dominated Convergence Theorem). Suppose $(X, \mathscr{A}, \mu)$ is a measure space, with $f, f_n: X \to [-\infty, +\infty]$, $\mathscr{A}$-measurable functions, and $g: X \to [-\infty, \infty]$ integrable. Suppose $f_n \to f$ and $|f_n| \leq g$ a.e. Then $\int f_n \to \int f$.

Here are the problems, adapted and condensed.

(Cohn, 2.4.2-2.4.4.) Prove the dominance convergence theorem in the following three steps.

  • (A) Show that the monotone convergence theorem still holds if the assumption $f_n$ are nonnegative is dropped, and instead the assumption $f_1$ is integrable is added.
  • (B) Show that if $(f_n)$ are measurable, non-increasing, $f_1$ integrable, then $\int f_n \to \int f$, with $f = \lim_n f_n$.
  • (C) Let $f, g, f_n$ be as in the statement of the dominated convergence theorem (see above for Cohn's version of it). Use the previous two parts, $p_n = \inf_{k \geq n} f_k$, $q_n = \sup_{k \geq n} f_k$ and the inequalities $p_n \leq f_n \leq q_n$ to furnish another proof of the dominated convergence theorem.

Here's what I did.

For part (A): Put $g_n = f_n - f_1$, for $n =1, 2, \dots$. Certainly, $g_n$ are nonnegative measurable functions, and moreover $g_n \uparrow (f - f_1)$, almost everywhere. Thus, the (usual) dominated convergence theorem gives $\int (f_n - f_1) = \int g_n \to \int (f - f_1)$, which by integrability of $f_1$ implies the result.

For part (B): We now have $f_1 \geq f_2 \geq \cdots$, and so $-f_1 \leq -f_2 \leq \cdots$ and $-f_1$ is integrable, and the conclusion follows from part (A).

For part (C): Let us first assume that $f_n \to f$ and $|f_n| \leq g$ everywhere. Then $\lim p_n = \lim q_n = f$, everywhere, and thus immediate consequences of (A) and (B) are $\int f = \int \lim p_n = \lim \int p_n$, and also $\int f = \lim \int q_n$, since $p_n$ and $q_n$ are monotone measurable sequences and $p_1$, $q_1$ are integrable since $|p_1|, |q_1| \leq g$. Now, the inequalities $p_n \leq f_n \leq q_n$ imply $\int p_n \leq \int f_n \leq \int q_n$, whence $$ \limsup \int f_n \leq \lim \int q_n = \int f = \lim \int p_n \leq \liminf \int f_n. $$ Thus $\int f = \lim \int f_n$, as needed.

For the almost everywhere result, let $N$ denote the set where either $f_n \not \to f$ or $|f_n| \leq g$ fails, which by assumption has measure zero. Now let us define $\tilde f_n = f_n \mathbb{1}_{N^c}$ and $\tilde f = f \mathbb{1}_{N^c}$ which satisfy $\tilde f_n \to \tilde f$ and also $|\tilde f_n| \leq g$, in which case our previous work implies $\int \tilde f_n \to \int \tilde f$, and hence $\int f_n - \int_N f_n \to \int f - \int_N f$, which implies the result since $\int_N f_n = \int_N f = 0$ for all $n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.