Proof of Riemann-Lebesgue lemma 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.
 A: Hint. Use the fact that every integrable function $f:[a,b]^2\to\mathbb R$ can be $L^1-$approximated by linear combinations of the form
$$
f_n=\sum_{j=1}^n c_n\chi_{R_n},
$$
where $R_j\subset [a,b]^2$, rectangle.
A: 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. $$\begin{align*}\mu\left(\left[i(n)-\frac{\varepsilon}{2^{n+2}},i(n)+\frac{\varepsilon}{2^{n+2}}\right]\right)&=\int\limits_{\mathbb{R}}1_{\left[i(n)-\frac{\varepsilon}{2^{n+2}},i(n)+\frac{\varepsilon}{2^{n+2}}\right]}=\int\limits_{\mathbb{R}}1_{\left[i(n)-\frac{\varepsilon}{2^{n+2}},i(n)+\frac{\varepsilon}{2^{n+2}}\right]}dx\\ &=\frac{\varepsilon}{2^{n+1}}\end{align*}$$
A: Write $f = g + h$, where $g$ is a linear combination of characteristic
functions of bounded rectangles and $||h|| < \epsilon$ in $L_1$.  The
statements are obvious with $f$ replaced by $g$ (as in the 1-dimensional
case), and the $L_1$ bound for $h$ ensures that the contribution of $h$
to the original integrals is sufficiently small.
The approximation exists for Lebesgue integrals because the definition
of an abstract measure-theory integral gives an approximation by
characteristic functions of Borel sets, and the definition of Lebesgue
measure gives approximations of Borel sets by unions of rectangles.
The approximation exists for Riemann integrals because the definition
of the integral gives suitable characteristic functions even more directly.
A previous version of this answer overcomplicated things for (a) using
Fubini's theorem.  It is important that the statements are for double
integrals, not for iterated integrals, since the iterated integrals
don't exist in general for Riemann integration and require Fubini's
theorem for Lebesgue integration. Using Fubini's theorem in the proof
only gave technical complications.
A: You can see Bochner, Chandrasekharan, "Fourier Transforms"
