The Riemann-Lebesgue Lemma

# The Riemann-Lebesgue Lemma

Recall from the Lebesgue Integrable Functions with Arbitrarily Small Integral Terms page that if $f \in L(I)$ then for all $\epsilon > 0$ there exists upper functions $u^*, v^* \in U(I)$ where $f = u^* - v^*$, $v^*$ is nonnegative almost everywhere on $I$, and $\displaystyle{\int_I v^*(x) \: dx < \epsilon}$.

We also saw that there exists $g \in L(I)$ and $s \in S(I)$ where $f = g + s$ and $\displaystyle{\int_I \mid g(x) \mid \: dx < \epsilon}$.

We will now use the latter result to prove an extremely important result called the Riemann-Lebesgue lemma.

 Lemma (The Riemann-Lebesgue Lemma): If $f \in L(I)$ and $\beta \in \mathbb{R}$ then $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = 0}$.
• Proof: We begin by showing that the Riemann-Lebesgue lemma holds for step functions. Let $[a, b] \subseteq I$ and define a function $f$ on $I$ by:
(1)
\begin{align} \quad f(t) = \left\{\begin{matrix} 1 & \mathrm{if} \: t \in [a, b]\\ 0 & \mathrm{if} \: t \in I \setminus [a, b] \end{matrix}\right. \end{align}
• Then we have that:
(2)
\begin{align} \quad \lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = \lim_{\alpha \to \infty} \int_a^b \sin (\alpha t + \beta) \: dt = \lim_{\alpha \to \infty} \left [ - \frac{\cos (\alpha t + \beta)}{\alpha} \right ]_a^b = \lim_{\alpha \to \infty} \frac{\cos (\alpha a + \beta) - \cos (\alpha b + \beta)}{\alpha} \end{align}
• However we see that:
(3)
\begin{align} \quad \lim_{\alpha \to \infty} \left (-\frac{2}{\alpha} \right ) \leq \lim_{\alpha \to \infty} - \biggr \lvert \frac{\cos (\alpha a + \beta) - \cos (\alpha b + \beta)}{\alpha} \biggr \rvert \leq \lim_{\alpha \to \infty} \frac{\cos (\alpha a + \beta) - \cos (\alpha b + \beta)}{\alpha} \leq \lim_{\alpha \to \infty} \biggr \lvert \frac{\cos (\alpha a + \beta) - \cos (\alpha b + \beta)}{\alpha} \biggr \rvert \leq \lim_{\alpha \to \infty} \frac{2}{\alpha} \end{align}
• So by the Squeeze theorem we see that indeed, $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = 0}$ for functions $f$ as defined above. However, every step function is the sum of functions of this type, so summing the limits of these functions shows that $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = 0}$ holds for every $f \in S(I)$.
• Now by the theorem mentioned at the top of this page, since $f \in L(I)$, for $\displaystyle{\epsilon_1 = \frac{\epsilon}{2} > 0}$ there exists functions $g \in L(I)$ and $s \in S(I)$ such that $f = g + s$ and where:
(4)
\begin{align} \quad \int_I \mid g(t) \mid \: dx = \int_I \mid f(t) - s(t) \mid \: dt < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
• And since we have already verified the Riemann-Lebesgue lemma to be true for step functions we have that $\displaystyle{\lim_{n \to \infty} \int_I s_n(t) \sin (\alpha t + \beta) \: dt = 0}$ so for $\displaystyle{\epsilon_2 = \frac{\epsilon}{2} > 0}$ there exists an $M \in \mathbb{R}$ such that if $\alpha \geq M$ then:
(5)
\begin{align} \quad \biggr \lvert \int_I s(t) \sin (\alpha t + \beta) \: dt \biggr \rvert < \epsilon_2 = \frac{\epsilon}{2} \quad (**) \end{align}
• Therefore for $\alpha \geq M$ we have that:
(6)
\begin{align} \quad \biggr \lvert \int_I f(t) \sin (\alpha t + \beta) \: dt \biggr \rvert & \leq \biggr \lvert \int_I [f(t) - s(t)] \sin (\alpha t + \beta) \: dt \biggr \rvert + \biggr \lvert \int_I s(t) \sin (\alpha t + \beta) \: dt \biggr \rvert \\ & \leq \int_I \mid f(t) - s(t) \mid \mid \sin (\alpha t + \beta) \mid \: dt + \biggr \lvert \int_I s(t) \sin (\alpha t + \beta) \: dt \biggr \rvert \\ & \leq \int_I \mid f(t) - s(t) \mid (1) \: dt + \biggr \lvert \int_I s(t) \sin (\alpha t + \beta) \: dt \biggr \rvert \\ & < \epsilon_1 + \epsilon_2 \\ & = \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ &= \epsilon \end{align}
• So for all $\epsilon > 0$ there exists an $M \in \mathbb{R}$ such that if $\alpha \geq M$ then $\displaystyle{\biggr \lvert \int_I f(t) \sin (\alpha t + \beta) \: dt \biggr \rvert < \epsilon}$ so for any $f \in L(I)$ and for all $\beta \in \mathbb{R}$, $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = 0}$. $\blacksquare$
 Corollary 1: If $f \in L(I)$ then: a) $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin \alpha t \: dt = 0}$. b) $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \cos \alpha t \: dt = 0}$.
• Proof of a) Set $\beta = 0$. Then by the Riemann-Lebesgue lemma we have that:
(7)
\begin{align} \quad \lim_{\alpha \to \infty} \int_I f(t) \sin \alpha t \: dt = 0 \quad \blacksquare \end{align}
• Proof of b) Set $\beta = \frac{\pi}{2}$. Once again, by the Riemann-Lebesgue lemma we have that:
(8)
\begin{align} \quad \lim_{\alpha \to \infty} \int_I f(t) \sin \left (\alpha t + \frac{\pi}{2} \right ) \: dt & = 0 \\ \quad \lim_{\alpha \to \infty} \int_I f(t) \cos \alpha t \: dt & = 0 \quad \blacksquare \end{align}