The Integrable Limit Theorem

One of the most fascinating explorations of integrable functions pertains its limiting conditions. When would limits of sequences of functions be themselves (Riemann-)integrable?

A sufficient condition is that if $f_n \to f$ uniformly, then $f$ must be integrable.

Theorem 1. Let $\{f_n\}$ be a sequence of integrable functions $f_n : [a, b] \to \mathbb R$ that converge point-wise to some function $f : [a, b] \to \mathbb R$ . If $f_n \to f$ uniformly, then $f$ is integrable, and

$\displaystyle \int_a^b f \equiv \int_a^b \lim_{n \to \infty} f_n = \lim_{n \to \infty} \int_a^b f_n.$

Proof. Fix $\epsilon > 0$ . Since $f_n \to f$ uniformly, for any $k_1 > 0$ , there exists $N \in \mathbb N$ such that

$j > N\quad \Rightarrow \quad \|f_j - f\|_\infty < k_1 \cdot \epsilon.$

In particular, the estimate works for $j = N+1$ . Since $f_j$ is integrable, for any $k_2 > 0$ , there exists a partition $P \subseteq [a, b]$ such that

$\displaystyle \sum_{i=1}^n (M_i(f_j, P) - m_i(f_j, P))\Delta x_i < k_2 \cdot \epsilon.$

For any $x \in [a, b]$ , unraveling the estimate yields

$\displaystyle f_j(x) - k_1 \cdot \epsilon< f(x) < f_j(x) + k_1 \cdot \epsilon,$

which yields the estimates

$\begin{aligned} M_i(f, P) &< M_i(f_j, P) + k_1 \cdot \epsilon,\\ m_i(f, P) &> m_i(f_j, P) - k_1 \cdot \epsilon. \end{aligned}$

By unraveling the definitions and combining the estimates,

$\begin{aligned} U(f, P) - L(f, P) &< U(f_j, P) - L(f_j, P) + 2k_1(b-a) \cdot \epsilon \\ &< k_2 \cdot \epsilon + 2k_1(b-a) \cdot \epsilon \\ &= (2k_1(b-a) + k_2) \cdot \epsilon.\end{aligned}$

Setting $k_1 = 1/(4(b-a)), k_2 = 1/2$ yields the desired integrability result. Furthermore, by the triangle inequality for integrals, for any $n > N$ ,

$\displaystyle \begin{aligned}\left| \int_a^b f_n - \int_a^b f \right| &= \left| \int_a^b (f_n - f )\right|\\ &\leq \int_a^b | f_n - f | \\ &\leq \int_a^b \| f_n - f \|_\infty \\ &\leq \| f_n - f \|_\infty \cdot (b-a) \\ &< k_1(b-a) \cdot \epsilon. \end{aligned}$

Setting $k_1 = 1/(b-a)$ yields the desired limit property

$\displaystyle \int_a^b f \equiv \int_a^b \lim_{n \to \infty} f_n = \lim_{n \to \infty} \int_a^b f_n.$

This is a classic example of valid instances when we can push the limit into the integral and vice versa. However, uniform convergence is typically too strong of a condition to be met, apart from some toy examples in standard tutorial problems.

With tools in measure theory, which we will explore a bit of next time, we can obtain a much more lenient condition. We will state this result, but not prove it, since its real utility shines forth in the context of measure theory.

Dominated Convergence Theorem. Let $\{f_n\}$ be a sequence of Lebesgue-integrable functions $f_n : [a, b] \to \mathbb R$ that converge point-wise to some function $f : [a, b] \to \mathbb R$ .

If there exists a nonnegative Lebesgue-integrable function $g : [a, b] \to \mathbb R$ such that $|f_n| \leq g$ and $|f| \leq g$ , then $f$ is Lebesgue-integrable, and

$\displaystyle \int_a^b f \equiv \int_a^b \lim_{n \to \infty} f_n = \lim_{n \to \infty} \int_a^b f_n.$

Proof. Omitted. For further discussion in measure theory.

Clearly, we cannot prove what we don’t even know—what even is a Lebesgue-integrable function? The reason this condition is more relaxed than Theorem 1 is because bounded functions on $[a, b]$ that are Riemann-integrable are Lebesgue-integrable too. Then the limit function $f$ will be Lebesgue integrable.

From an applied perspective, probability theorists model random phenomena using random variables, which are defined to be Lebesgue-integrable (more technically, Lebesgue-measurable) functions. From probability theory arises all sorts of applications across various STEM fields.

For now, that’s it for undergraduate real analysis!

—Joel Kindiak, 24 Jan 25, 1452H

KindiakMath

The Integrable Limit Theorem

Response

Leave a reply to Infinite Series Revisited – KindiakMath Cancel reply

The Integrable Limit Theorem

Share this:

Response

Leave a reply to Infinite Series Revisited – KindiakMath Cancel reply