Riemann-Lebesgue lemma for periodic functions Suppose $f$ is a periodic function of period $T$ and $g$ is integrable on $[a,b]$. How do I prove that $$\lim_{t\to\infty} \int_a^b f(tx)g(x) \, dx = \frac 1 T \int_0^T f(x) \, dx \int_a^b g(x)\,dx $$
The problem didn't indicate that $f$ is continuous.
 A: Suppose $f$ is a periodic function of period $T$ and $g$ is integrable on $[a,b]$. We will show that $$\lim_{t\to\infty} \int_a^b f(tx)g(x) \, dx = \frac 1 T \int_0^T f(x) \, dx \int_a^b g(x)\,dx. $$
Proof: We are given $f$ is integrable. Denote $f_{avg}=\displaystyle{\frac{1}{T}\int_0^Tf(x)\,dx}$ as the average of $f$ over $[0,T]$. Denote $F$ as the anti-derivative of $f$. Then we have 
\begin{align*}
F(x) &= \displaystyle {\bigg(\frac{1}{T}\int_0^Tf(t)\,dt\bigg)}x+g(x),\\
& \text{where $T>0$ is the period of $f$ and $g:\mathbb{R}\to\mathbb{R}$ is a $T$-periodic function.} \\ &\text{Lemma (1)}
\end{align*}
To prove this lemma, we may note that for any periodic function $f$, $f(t+T)=f(t)$ holds true for any $t\in\mathbb{R}$, so it follows that $F(x+t)-F(x)=\int_0^T f(t)\,dt$ for all $x\in\mathbb{R}$. We consider the function $h(x)=\displaystyle{\bigg(\frac{1}{T}\int_0^T f(t)\,dt}\bigg)x,$ we have $h(x+T)-h(x)=\int_0^Tf(t)\,dt$; and with our previous result $F(x+T)-h(x+T)=F(x)-h(x).$ This is the function defined by $g(x)=F(x)-h(x),\,x\in\mathbb{R}$, and is periodic of period $T$. This confirms lemma (1). Now using our lemma, we define our function $F(x)=f_{avg}\cdot\,x+h(x)$ for $x\in[a,b]$ where $h$ is continuous and has period $T$. 
We will write
\begin{align*}
& t\int_0^T\,g(x)f(tx)\,dx = \int_0^T\,g(x)F'(tx)\,dx = g(x)F(tx)\bigg|_{0}^{T}-\int_0^T\,g'(x)F(tx)\,dx=\\
&=g(T)F(tT)-\int_0^T\,g'(x)F(tx)\,dx = g(T)(f_{avg}tT+h(tT))-\\&-tf_{avg}\int_0^T\,xg'(x)\,dx-\int_0^T\,g'(x)h(tx)\,dx = tf_{avg}Tg(T)-\\&-tf_{avg}Tg(T)+tf_{avg}Tg(T)+tf_{avg}\int_0^T\,g(x)\,dx-\int_0^T\,g'(x)h(tx)\,dx = \\
&= tf_{avg}\int_0^T\,g(x)\,dx-\int_0^T\,g'(x)h(tx)\,dx.
\end{align*} 
That is equivalent to 
\begin{align*}
t\bigg(\displaystyle{\int_0^T\,g(x)f(tx)\,dx-\frac{1}{T}\int_0^T\,f(x)\,dx\int_0^T\,g(x)\,dx\bigg)= -\int_0^T\,g'(x)h(tx)\,dx.} (2)
\end{align*}
Then by the Riemann-Lebesgue lemma we have 
\begin{align*}
& \displaystyle{\lim_{t\to\infty}\int_0^T\,g'(x)h(tx)\,dx = \frac{1}{T}\int_0^Th(x)\,dx\int_0^T\,g'(x)\,dx} = \\
&= \frac{1}{T}(g(T)-g(0))\int_0^T\,h(x)\,dx\int_0^T\,g'(x)\,dx = \\
&=\frac{1}{T}(g(T)-g(0))\bigg(\int_{0}^{T}F(x)\,dx-F(T)\bigg)
\end{align*}
and using (2) we have our result. $\blacksquare$
Source: http://emis.ams.org/journals/AUA/acta8/Andrica-Piticari.pdf
