TITLE: Proof of Riemann-Lebesgue lemma
QUESTION [7 upvotes]: I read a book, and this mention to the following lemma of Rieman-Lebesgue type. 
Lemma. Let $-\infty<a<b<\infty$ and $f(x,y):[a,b]^2\to\mathbb R$ be an integrable and nonnegative function. Then we the followings.
a) $\lim\limits_{r\to\infty}\int\limits_{[a,b]^2}f(x,y)\sin\left(\pi r x\right) \, dx \, dy=0$
b) $\lim\limits_{r\to\infty}\int\limits_{[a,b]^2}f(x,y)\sin\left(\pi r x\right)\sin(\pi ry) \, dx \, dy=0$.
I try to find a proof, my idea is to use step function, but I fail. Does anyone know some proof for this lemma? Thank you.

REPLY [6 votes]: Let $f(x,y)$ be R-integrable in $\mathbb{R}^2$-space, $L^1(\mathbb{R^2})$ function which returns only numbers. I am going to prove that:
$$\lim_{|\omega_x +i\omega_y| \to +\infty } \, \int\limits_{\mathbb{R}^2} f(x,y)\ e^{- i (x \omega_x+y \omega_y)}\,dxdy=0$$
$(\omega_x,\omega_y)\in \mathbb{R}^2$ 
Let's start from:
$$\int\limits_{\mathbb{R}^2} f(x,y)\ e^{- i (x \omega_x+y \omega_y)}\,dxdy=\int\limits_{\mathbb{R}^2} f(x,y)\ e^{- i (x \alpha+y \beta)\omega}\,dxdy=\int\limits_{\mathbb{R}^2} g(x_2,y_2)\ e^{- i x_2\omega}\,dx_2dy_2=\int\limits_{\mathbb{R}} h(x_2)\ e^{- i x_2\omega}\,dx_2$$
Where: $\omega=|\omega_x+i\omega_y|$ , $\alpha=\frac{\omega_x}{\omega}$ ,$\beta=\frac{\omega_y}{\omega} \implies |\alpha|^2+|\beta|^2=1$, without loss of generality I can assume that $|\omega_x|\geqslant |\omega_y|$, so $|\alpha|^2\geqslant\frac{1}{2}$.
$\left[\begin{array}\
y_2\\
x_2
\end{array}\right]=\left[\begin{array}\
1 & 0\\
\beta & \alpha
\end{array}\right]\cdot\left[\begin{array}\
y\\
x
\end{array}\right]\implies$$\left|\frac{D(x,y)}{D(x_2,y_2)}\right|$$=\frac{1}{|\alpha|}$ , $g(x_2,y_2)=\frac{1}{|\alpha|}f\left(\frac{x_2-\beta y_2}{\alpha},y_2\right)$
$h(x_2)=\int_{-\infty}^{+\infty} g(x_2,y_2)\,dy_2$
We have here bijection $(x,y)$ into $(x_2,y_2)$ and $\mathbb{R}^2$ as area of integrating for $f(x,y)$ what determines that equivalent area of integrating for $g(x_2,y_2)$ is also $\mathbb{R}^2$.
Note:
$$\int\limits_{\mathbb{R}^2} |f(x,y)|\,dxdy=\int\limits_{\mathbb{R}^2} |g(x_2,y_2)|\,dx_2dy_2\geqslant\int_{-\infty}^{+\infty}\left|\int_{-\infty}^{+\infty} g(x_2,y_2)\,dy_2\right|dx_2=\int\limits_{\mathbb{R}} |h(x_2)|\,dx_2$$ 
so $h$ is $L^1(\mathbb{R})$ function.If $h$ is not integrable then it is against assumption about integrability of $f$. Let's introduce $\widetilde{h}$ as: $\widetilde{h}(x,\Delta,n)=\sum_{j=-n}^{n}h(x_j^{*})$$\Pi$$\left(\frac{x-j\cdot\Delta}{\Delta}\right)$, where $n\in\mathbb{Z}_{+}$, $\Delta > 0$ 
$$x_j^{*}:|h(x_j^{*})|=\min\left\{|h(x)|:j\Delta-\frac{\Delta}{2}\leqslant x\leqslant j\Delta+\frac{\Delta}{2}\right\}\ \tag{1}$$ 
If $\widetilde{h}(x,\Delta)=\widetilde{h}(x,\Delta,+\infty)$ then:
$\int\limits_{\mathbb{R}} |\widetilde{h}(x,\Delta)|\,dx\leqslant \int\limits_{\mathbb{R}} |h(x)|\,dx<+\infty$, what means belonging to $L^1(\mathbb{R})$. 
Let $l:\mathbb{R}_{+}\to \mathbb{R}_{+}\cup \{0\}$ and $l(\Delta)=\lim\limits_{\omega \to +\infty } \, \left|\int\limits_{\mathbb{R}} \widetilde{h}(x,\Delta)e^{-i\omega x}\,dx\right|$ so:
 $$\begin{align*} \lim_{\omega \to +\infty } \, \int\limits_{\mathbb{R}} \widetilde{h}(x,\Delta)e^{-i\omega x}\,dx  &=  \lim\limits_{\omega\to +\infty}\lim_{n \to +\infty }\sum_{j=-n}^{n}h(x_j^{*})\int_{j\Delta-\frac{\Delta}{2}}^{j\Delta+\frac{\Delta}{2}}e^{-i \omega x}\,dx\\ &= \lim\limits_{\omega\to +\infty}\lim_{n \to +\infty }\sum_{j=-n}^{n}h(x_j^{*})\left[\frac{\Delta e^{-i \omega x}}{-i \omega \Delta}\right]{j\Delta+\frac{\Delta}{2} \atop j\Delta-\frac{\Delta}{2}}\\
l(\Delta)&\leqslant \lim\limits_{\omega\to +\infty}\lim_{n \to +\infty }\sum_{j=-n}^{n}|h(x_j^{*})|\frac{2\Delta}{\omega\Delta}\\ &\leqslant \lim_{\omega \to +\infty }\frac{2\int_{-\infty}^{+\infty}|h(x)|\,dx}{\omega\Delta}=\frac{1}{\Delta}\lim_{\omega \to +\infty }\frac{2\int_{-\infty}^{+\infty}|h(x)|\,dx}{\omega}=0 \end{align*}$$
what implies $l(\Delta)=0$ and $\lim\limits_{\Delta\to 0^{+}} l(\Delta)=0$. By the triangle inequality we obtain:
$$\begin{align*}\lim\limits_{\Delta\to 0^{+}}\lim_{\omega \to +\infty } \, \left|\int\limits_{\mathbb{R}} h(x)e^{-i\omega x}\,dx\right|&\leqslant \lim\limits_{\Delta\to 0^{+}}\lim_{\omega \to +\infty } \, \left|\int\limits_{\mathbb{R}} (h(x)-\widetilde{h}(x,\Delta))e^{-i\omega x}\,dx\right|+\lim_{\Delta \to 0^{+} } l(\Delta)\\
&\leqslant \lim_{\Delta \to 0^{+} }\lim\limits_{\omega\to +\infty} \, \int\limits_{\mathbb{R}} |h(x)-\widetilde{h}(x,\Delta))|\,dx\end{align*}$$
So:
$$\begin{align*}\lim_{\omega \to +\infty } \, \left|\int\limits_{\mathbb{R}} h(x)e^{-i\omega x}\,dx\right|\leqslant \lim_{\Delta \to 0^{+} } \, \int\limits_{\mathbb{R}} |h(x)-\widetilde{h}(x,\Delta))|\,dx&=\int\limits_{\mathbb{R}} |h(x)|\,dx-\lim_{\Delta \to 0^{+} } \, \int\limits_{\mathbb{R}} |\widetilde{h}(x,\Delta))|\,dx\\ &=0 \end{align*}$$
Fact that $|h(x)-\widetilde{h}(x,\Delta))|=|h(x)|-|\widetilde{h}(x,\Delta)|$ is related with $(1)$ and  Lebesgue's criterion for Riemann integrability which indicates that for any R-integrable function integrating discontinuities area can be skipped (requirement of continuity almost everywhere). I've proved that:
$$\lim_{|\omega_x +i\omega_y| \to +\infty } \, \int\limits_{\mathbb{R}^2} f(x,y)\ e^{- i (x \omega_x+y \omega_y)}\,dxdy=\lim_{\omega \to +\infty } \, \int\limits_{\mathbb{R}} h(x)e^{-i\omega x}\,dx=0$$
If function $f$ is not R-integrable , but exists R-integrable,$L^1(\mathbb{R}^2)$ function which is equal to it almost everywhere then we can obtain this same result here by replacing $f$ with its estimate or just using L-integrating. Lebesgue integral of non-negative measurable function $g:\mathbb{R}^n\to [0,+\infty)$ is equal to: $\lim\limits_{\Delta\to 0^{+}}\lim\limits_{n\to +\infty}\sum_{i=1}^n\Delta$$\mu$$(A_i)$, where $A_i=\{x\in\text{area of integrating}:g(x)>i\Delta\}$.
Curiosity:
$f_1(x)=($$1_{\mathbb{Q}}$$2-1)\frac{1}{1+x^2} \implies \lim\limits_{\omega \to +\infty } \, \int\limits_{\mathbb{R}} f_1(x)e^{-i\omega x}\,dx\neq 0 \wedge \lim\limits_{\omega \to +\infty } \, \int\limits_{\mathbb{R}} f_1(x)e^{-i\omega x}=0$
The second result is more authoritative because $\mu(\mathbb{Q})=0$. Coverage of any countable set $X$ can be expressed as $\bigcup\limits_{n=1}^{+\infty}\left[i(n)-\frac{\varepsilon}{2^{n+2}},i(n)+\frac{\varepsilon}{2^{n+2}}\right]$, where $\varepsilon > 0$ and $i\colon \mathbb{Z}_+ \to X$ is a surjection. $$\begin{align*}\mu\left(\left[i(n)-\frac{\varepsilon}{2^{n+2}},i(n)+\frac{\varepsilon}{2^{n+2}}\right]\right)&=\int\limits_{\mathbb{R}}1_{\left[i(n)-\frac{\varepsilon}{2^{n+2}},i(n)+\frac{\varepsilon}{2^{n+2}}\right]}=\int\limits_{\mathbb{R}}1_{\left[i(n)-\frac{\varepsilon}{2^{n+2}},i(n)+\frac{\varepsilon}{2^{n+2}}\right]}dx\\ &=\frac{\varepsilon}{2^{n+1}}\end{align*}$$