Crit. for an Imp. Riemann Integrable Function to be Lebesgue Integrable

# Criterion for an Improper Riemann Integrable Function to be Lebesgue Integrable

Recall that a function $f$ may be improper Riemann integrable on an unbounded interval, say $I = [a, \infty)$ (where $a \in \mathbb{R}$) but might not be Lebesgue integrable on $I$.

We will now establish a nice theorem which gives us a condition for which an improper Riemann integrable function $f$ on $I = [a, \infty)$ is also Lebesgue integrable on $I$.

 Theorem 1: $f$ be an improper Riemann integrable function on the interval $[a, b]$ for all $b > a$ and suppose that there exists an $M \in \mathbb{R}$, $M > 0$ such that the Riemann integral of $\mid f \mid$ is uniformly bounded by $M$ for all $b > a$, i.e., $\displaystyle{\int_a^b \mid f(x) \mid \: dx \leq M}$ for all $b > a$. Then: a) $\mid f \mid$ is improper Riemann integrable on $[a, \infty)$. b) $f$ is improper Riemann integrable on $[a, \infty)$. c) $f$ is Lebesgue integrable on $[a, \infty)$. d) $\displaystyle{\underbrace{\int_a^{\infty} f(x) \: dx}_{\mathrm{Lebesgue \: Integral}} = \lim_{b \to \infty} \int_a^b f(x) \: dx = \underbrace{\int_a^{\infty} f(x) \: dx}_{\mathrm{Improper \: Riemann \: Integral}}}$.
• Proof of a) To show that $\mid f \mid$ is improper Riemann integrable on $[a, \infty)$ we must show that $\displaystyle{\lim_{b \to \infty} \int_a^b \mid f(x) \mid \: dx}$ exists. Define a function $F : [a, \infty) \to \mathbb{R}$ for all $b \in [a, \infty)$ as a Riemann integral function:
(1)
\begin{align} \quad F(b) = \int_a^b \mid f(x) \mid \: dx \end{align}
• Note that $\mid f \mid$ is a strictly nonnegative function. Furthermore, we are given 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 $b > a$. Therefore $F(x)$ is an increasing function that is bounded above. So $\displaystyle{\lim_{b \to \infty} F(b) = \lim_{b \to \infty} \int_a^b \mid f(x) \mid \: dx}$ exists which shows that $\mid f \mid$ is improper Riemann integrable on $[a, \infty)$. $\blacksquare$
• Proof of b) Define a new function $G : [a, \infty) \to \mathbb{R}$ for all $x \in [a, \infty)$ by:
(2)
\begin{align} \quad G(x) = \mid f(x) \mid - f(x) \end{align}
• Note that $f(x) \leq \mid f(x) \mid$ for all $x \in [a, \infty)$ and so $G$ is a nonnegative function. Define a new function, $F^* : [a, \infty) \to \mathbb{R}$ for all $b \in [a, \infty)$ by:
(3)
\begin{align} \quad F^*(b) = \int_a^b G(x) \: dx \end{align}
• Then since $G$ is a nonnegative function we have that $F^*$ is an increasing function on $[a, \infty)$. Moreover, for all $b > a$ we have that:
(4)
\begin{align} \quad 0 \leq F^*(x) = \int_a^b G(x) \: dx \leq \int_a^b [\mid f(x) \mid - f(x) \mid] \leq \int_a^b 2 \mid f(x) \mid \: dx = 2 \int_a^b \mid f(x) \mid \: dx \leq 2M \end{align}
• So $F^*(b)$ is an increasing function that is bounded above. This implies that $\displaystyle{\lim_{b \to \infty} F^*(b) = \lim_{b \to \infty} \int_a^b G(x) \: dx}$ exists, i.e., the following Riemann integral exists:
(5)
\begin{align} \quad \lim_{b \to \infty} \int_a^b G(x) \: dx = \lim_{b \to \infty} \int_a^b [\mid f(x) \mid - f(x)] \: dx \end{align}
• So $G = \mid f \mid - f$ is improper Riemann integrable on $[a, \infty)$. From part (a) we also have that $\mid f \mid$ is improper Riemann integrable on $[a, \infty)$. So the difference $\mid f \mid - (\mid f \mid - f) = f$ is improper Riemann integrable on $[a, \infty)$. $\blacksquare$
• Proof of c) We know that $f$ is Riemann integrable on each closed and bounded interval $[a, b]$ for $b > a$. So $f$ is Lebesgue integrable on $[a, b]$ for all $b > a$. Furthermore, we're given that $\displaystyle{\int_a^b \mid f(x) \mid \: dx \leq M}$ for all $b > a$. Therefore by the theorem presented on the Lebesgue Integrals on Unbounded Intervals page, $f$ must be Lebesgue integrable on $[a, \infty)$.
• Proof of d) From the theorem referenced above, we also have that:
(6)
\begin{align} \quad \underbrace{\int_a^{\infty} f(x) \: dx}_{\mathrm{Lebesgue \: Integral}} = \lim_{b \to \infty} \int_a^b f(x) \: dx = \underbrace{\int_a^{\infty} f(x) \: dx}_{\mathrm{Improper \: Riemann \: Integral}} \quad \blacksquare \end{align}