# Help with a Lebesgue integration problem.

The question is the following,

Let $f:\Bbb{R}\rightarrow \Bbb{R}$ be a Lebesgue integrable function. Show that $\lim\limits_{t \to \infty } \int_\Bbb{R} f(x)\cos (xt) \, dx = \lim\limits_{t \to \infty } \int_\Bbb{R} f(x)\sin (xt) \, dx = 0$

My unsuccessful attempt $\lim\limits_{t \to \infty } \int_\Bbb{R} f(x)\cos (xt) \, dx = \lim\limits_{t \to \infty } \int_\Bbb{R} f(\frac{u}{t})\frac{\cos (u)}{t} \, dt$. Since $f$ is Lebesgue measurable, then $f(\frac{u}{t})$ is finite a.e., so $f(\frac{u}{t}) \le a$ for some constant $a$, so $\left| f(\frac{u}{t})\frac{\cos (u)}{t} \right| \le \left| \frac{a\cos (u)}{t} \right| \le \left| a \right|$.

However, $\left| a \right|$ is not Lebesgue integrable on $\Bbb{R}$, and I still cannot exchange the limit and integration by dominated convergence theorem, so I got stuck.

Thank you!

• First show the claim for $C_c^\infty$ functions (by partial integration), then use density. What you want to show is a variant of the Riemann Lebesgue Lemma. – PhoemueX Jul 24 '15 at 5:56

First we consider it on $[a,b]$. Since $f$ is Lebesgue integrable, $|f|$ is too. Let $A=\{x:x\in \Bbb{R},|f(x)|> n\}$. Then given $\epsilon>0,$ there is $N$ such that for any $n>N$ $$\int_{A}|f|dm<\epsilon/4\tag{1}$$ By Lusin's theorem, there is a continuous function $g$ on $[a,b]$ that $$m(\{x:f(x)\ne g(x), \:x\in[a,b]\})<\epsilon/2MN\tag{2}$$ Note $|g|<M$ on $[a,b]$. Then \begin{align} \left|\int_{a}^{b}f(x)\cos(xt) \, dx\right|&=\left|\int_{a}^{b}(f(x)-g(x))\cos(xt) \, dx+\int_{a}^{b}g(x)\cos(xt) \, dx\right| \\ &=\left|\int_{\{x:f\ne g, \:x\in[a,b]\}\cap(A\cup A^c)}(f(x)-g(x))\cos(xt) \, dx +\int_{a}^{b}g(x)\cos(xt) \, dx\right| \\ &<\int_{\{x:f\ne g, \:x\in[a,b]\}\cap A}|f(x)| \, dx+\int_{\{x:f\ne g, \:x\in[a,b]\}\cap A^c}|f(x)| \, dx \\&\quad+\int_{\{x:f\ne g, \:x\in[a,b]\}}|g(x)| \, dx+\left|\int_{a}^{b}g(x)\cos(xt) \, dx\right| \\&\hspace{110 mm}\text{by (1),(2)} \\ &<\epsilon /4+\frac{\epsilon N}{2MN}+\frac{\epsilon M}{2MN}+\left|\int_{a}^{b}g(x)\cos(xt) \, dx\right| \\ &<\epsilon/4+\epsilon/2+\epsilon/4 +\left|\int_{a}^{b}g(x)\cos(xt) \, dx\right| \\ &=\epsilon +\left|\int_{a}^{b}g(x)\cos(xt) \, dx\right|\\ \end{align} Last step assumes $M>1,N>2$. So $\int_{a}^{b}f(x)\cos(xt) \, dx\to0$ if $\int_{a}^{b}g(x)\cos(xt) \, dx\to0$. Thus we only prove it for continuous function.
If $f$ is continuous on $[a,b]$, it is Riemann integrable. So there exists a partition $P$ defined by the points $a=x_0 < x_1 < \ldots < x_n = b$ such that $$0 \leqslant \int_a^b f(x) dx - \sum_{i=1}^{n}m_i(x_i - x_{i-1}) < \epsilon\tag{3}$$ where $m_i = \inf_{x \in [x_{i-1}, x_i]} f(x)$. In other words, $f$ can be approximated from below by a step function.
By $(3)$, there is $$0 \leqslant \sum_{i=1}^{n}\int_{x_{i-1}}^{x_i} (f(x) - m_i) dx < \epsilon$$ So \begin{aligned} \left|\sum_{i=1}^{n}\int_{x_{i-1}}^{x_i} (f(x) - m_i) \cos(xt) dx\right| & \leqslant \sum_{i=1}^{n}\int_{x_{i-1}}^{x_i} |f(x) - m_i| dx \\ &= \sum_{i=1}^{n}\int_{x_{i-1}}^{x_i} (f(x) - m_i) dx \\ &< \epsilon \hspace{95 mm}\text{(4)} \end{aligned} There exists $T$, such that for any $t>T$ $$\left|\int_{x_{i-1}}^{x_i} m_i \cos(xt) dx\right| \leqslant \frac{2|m_i|}{t}\leqslant \frac{2M'}{t}<\epsilon/n\tag{5}$$ where $M'=\max{\{m_i,\:1 \leqslant i\leqslant n\}}$. So by $(4),(5)$, for $t>T$ \begin{aligned} \left|\int_{a}^{b}f(x)\cos(xt)dx\right| &= \left|\sum_{i=1}^{n}\int_{x_{i-1}}^{x_i} f(x)\cos(xt) dx\right| \\ &\leqslant \left|\sum_{i=1}^{n}\int_{x_{i-1}}^{x_i} (f(x) - m_i)\cos(xt) dx\right| + \left|\sum_{i=1}^{n}\int_{x_{i-1}}^{x_i} m_i \cos(xt) dx\right| \\ &< 2\epsilon \end{aligned}
Finally letting $a\to-\infty,b\to\infty$, we can prove it for $\Bbb{R}$.