Lebesgue Integrals on Unbounded Intervals

# Lebesgue Integrals on Unbounded Intervals

Recall from the Levi's Monotone Convergence Theorems page that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions on $I$ that are increasing almost everywhere on $I$ and such that $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ exists, then $(f_n(x))_{n=1}^{\infty}$ converges to a limit function $f$ which is Lebesgue integrable on $I$ and:

(1)
\begin{align} \quad \int_I f(x) \: dx = \lim_{n \to \infty} \int_I f_n(x) \: dx \end{align}

Also recall from the Lebesgue's Dominated Convergence Theorem page that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions on $I$ that converge to a limit function $f$ almost everywhere on $I$ and there exists a function $g$ that is Lebesgue integrable on $I$ and such that $\mid f_n(x) \mid \leq g$ almost everywhere on $I$ and for all $n \in \mathbb{N}$, then $f$ is Lebesgue integrable on $I$ and also:

(2)
\begin{align} \quad \int_I f(x) \: dx = \lim_{n \to \infty} \int_I f_n(x) \: dx \end{align}

Now consider a function $f$ defined on $I$ where $I$ is an unbounded interval, say $I$ is either $[a, \infty)$, $(-\infty, a]$, or $(-\infty, \infty)$ where $a \in \mathbb{R}$.

Then of course, $f$ may or may not be Lebesgue integrable on $I$. The following theorem gives us a nice condition for when $f$ is. We will use both Levi's monotone convergence theorem for Lebesgue integrals AND Lebesgue's dominated convergence theorem in the proof below.

 Theorem 1: Let $f$ be a function defined on the interval $I = [a, \infty)$ and suppose that $f$ is Lebesgue integrable on $[a, b]$ for all $b > a$. Furthermore, suppose that there exists an $M \in \mathbb{R}$, $M > 0$ such that $\displaystyle{\int_a^b \mid f(x) \mid \: dx \leq M}$ for all $b > a$. Then $f$ is Lebesgue integrable on $I = [a, \infty)$ and $\displaystyle{\int_I f(x) \: dx = \lim_{b \to \infty} \int_a^b f(x) \: dx}$.

Theorem 1 can be rewritten and proven true if $I = (-\infty, a]$, and we can use the result above and this modified result to show that this theorem is also true if $I = (-\infty, \infty)$.

• Proof: Let $(a_n)_{n=1}^{\infty}$ be a strictly increasing sequence of real numbers contained in the interval $I = [a, \infty)$. Define a sequence of functions $(f_n(x))_{n=1}^{\infty}$ for all $n \in \mathbb{N}$ as follows:
(3)
\begin{align} \quad f_n(x) = \left\{\begin{matrix} f(x) & \mathrm{if} \: x \in [a, a_n]\\ 0 & \mathrm{if} \: x \in (a_n, \infty) \end{matrix}\right. \end{align}
• Then each function $f_n$ is Lebesgue integrable on $I = [a, \infty)$. To see this, we note that $f = f_n$ on $[a, a_n]$ and since $f$ is Lebesgue integrable on $[a, b]$ for all $b > a$, then for $b = a_n$ we see that $f$ is Lebesgue integrable on $[a, a_n]$. Moreover, $0$ is Lebesgue integrable on $(a_n, \infty)$. So, by one of the theorems on the Additivity of Lebesgue Integrals on Subintervals of General Intervals page we have that each $f_n$ is Lebesgue integrable on $I = [a, \infty)$.
• Furthermore, it's not hard to see that the sequence of functions $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$ on $I = [a, \infty)$. Moreover, the sequence of functions $(\mid f_n(x) \mid)_{n=1}^{\infty}$ also converges to $\mid f(x) \mid$ on $I = [a, \infty)$. The sequence $(\mid f_n(x) \mid)_{n=1}^{\infty}$ is also an increasing sequence.
• We are given that $\displaystyle{\int_a^b \mid f(x) \mid \: dx \leq M}$, and so for each $n \in \mathbb{N}$ we have that:
(4)
\begin{align} \quad \int_I \mid f_n(x) \mid \: dx = \int_a^{\infty} \mid f_n(x) \mid \: dx = \int_a^{a_n} \mid f_n(x) \mid \: dx + \int_{a_n}^{\infty} 0 \: dx = \int_a^{a_n} \mid f_n(x) \mid \: dx \leq M \end{align}
• Therefore $\displaystyle{\lim_{n \to \infty} \int_I \mid f_n(x) \mid \: dx}$ exists. So we can apply Levi's Monotone Convergence Theorems which tells us that the limit function $\mid f \mid$ is Lebesgue integrable on $I = [a, \infty)$.
• Note that for all $n \in \mathbb{N}$ that then $\mid f_n(x) \mid \leq \mid f(x) \mid$. So $\mid f(x) \mid$ is Lebesgue integrable and dominates the sequence $(f_n(x))_{n=1}^{\infty}$. So, by Lebesgue's Dominated Convergence Theorem we have that $f$ is Lebesgue integrable on $I = [a, \infty)$ and:
(5)
\begin{align} \quad \int_I f(x) \: dx = \int_a^{\infty} f(x) \: dx = \lim_{n \to \infty} \int_a^{a_n} f(x) \: dx = \lim_{b \to \infty} \int_a^b f(x) \: dx \quad \blacksquare \end{align}