Arzelà's theorem

Here we list two useful theorems without proofs. If you learn the theory of measure and Lebesgue integral, you will encounter more general versions of these theorems. However, the following theorems do not require the Lebesgue integral but hold for the Riemann integral.

Theorem (Arzelà's convergence theorem)

Suppose that the sequence of continuous functions

$$f_n=f_n(x)(n=1,2,\cdots )$$

is uniformly bounded on the closed bounded interval $I=[a,b]$. That is, there exists an $M>0$ such that

$$|f_n(x)|\le M(x\in I,n=1,2,\cdots ).$$

In addition, suppose that $\{f_n\}$ converges to a continuous function $f_0=f_0(x)$ at each $x\in I$. That is,

$$\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)=f_0(x)(x\in I).$$

Then, we have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_a^b{f}_n(x)dx={\int }_a^b{f}_0(x)dx.$$

In other words, the limit as $n\to \mathrm{\infty }$ and the integral are ``commutative'' (i.e., the order of limit and integration can be swapped).

Perhaps it is clearer if we write

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_a^b{f}_n(x)dx={\int }_a^b\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)dx.$$

This theorem is a special case of Lebesgue's Bounded Convergence Theorem.

Theorem (Lebesgue's bounded convergence theorem)

Let $\{f_n\}$ be a sequence of uniformly bounded measurable functions that converges point-wise to a measurable function $f_0$ on a measure space $(S,\mathrm{\Sigma },\mu )$. Then,

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_S{f}_n(x)\mu (dx)={\int }_S{f}_0(x)\mu (dx).$$

Theorem (Extended Arzelà's convergence theorem)

Let the sequence of functions $\{f_n\}$ and the function $f_0$ be defined on $\mathbb{R}=(-\mathrm{\infty },\mathrm{\infty })$. Suppose the following conditions:

1. $f_n(n=1,2,\cdots )$ and $f_0$ are continuous everywhere except for finitely many points.
2. There exists a (piece-wise continuous) function $g(x)$ such that $$|f_n(x)|\le g(x)(x\in \mathbb{R})$$ and $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}g(x)dx<+\mathrm{\infty }.$$
3. $$f_n(x)\to f_0(x)(n\to \mathrm{\infty })$$ holds everywhere except for finitely many points.

Then, we have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}_n(x)dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}_0(x)dx.$$

This is a special case of Lebesgue's Dominated Convergence Theorem.

Theorem (Lebesgue's Dominated Convergence Theorem)

  Let $\{f_n\}$ be a sequence of measurable functions on a measure space $(S,\mathrm{\Sigma },\mu )$. Suppose that $\{f_n\}$ point-wise converges to a function $f_0$ and is dominated by some Lebesgue-integrable function $g(x)$:

$$|f_n(x)|\le g(x)$$

for all $n\in \mathbb{N}$ and all $x\in S$. Then $f_0$ is Lebesgue-integrable and

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_S{f}_n(x)\mu (dx)={\int }_S{f}_0(x)\mu (dx).$$