A measurable function Suppose $f\geq 0$ be a measurable function such that
$$\int_Bf\leq \frac{m(B)}{1+m(B)}$$ for any ball.
Then show that 
$$\int_{\mathbb{R}^n}f(kx)g(x)\to 0$$
as $k\to \infty$ for any $g$ integrable in $\mathbb{R}^n$.
 A: This is kind of a Riemann-Lebesgue Lemma, but I'm not sure if we can obtain a proof just from that. 
Anyway, let us first note that if we prove the assertion for $g\geq0$ we are done, since
$$
\left|\int_{\mathbb{R}^n}f(kx)g(x)\,dm\right|\leq\int_{\mathbb{R}^n}f(kx)|g(x)|\,dm.
$$
The hypothesis on $f$ implies that $f$ is integrable and $\int f\,dm\leq1$. Indeed, let $B_n$ be the ball of radius $n$ centered at the origin. Then, by Monotone Convergence, $$
\int f\,dm=\lim_n\int f\; 1_{B_n}\,dm=\lim_n\int_{B_n} f\,dm\leq 1.
$$ 
Knowing that $f$ is integrable, almost every point in $\mathbb{R}^n$ is a  Lebesgue point, so  $f\leq1$ a.e. by the inequality
$$
\frac1{m(B)}\int_Bf\,dm\leq\frac1{1+m(B)}.
$$
The integrability of $g$ implies that 
$$
\lim_{K\to\infty}\int_{g>K}g=0.
$$
Given $\varepsilon>0$, we choose $K$ such that $\int_{g>K}g\,dm<\varepsilon$. Then, writing $g_K$ for the function $g\;1_{g\leq K}$,
\begin{eqnarray}
\int f(kx)g(x)\,dm&=&\int_{g\leq K}f(kx)g(x)\,dm +\int_{g>K}f(kx)g(x)\,dm\\
&\leq&\int f(kx)g_K(x)\,dm +\varepsilon\\
&=&\frac1{k^n}\int f(t) g_K(t/k)\,dm+\varepsilon\\
&\leq&\frac{K}{k^n}+\varepsilon.
\end{eqnarray}
Thus
$$
\limsup_{k\to\infty}\int f(kx)g(x)\,dm\leq\varepsilon.
$$
As $\varepsilon$ was arbitrary, we conclude that the $\limsup$ is zero, and so the limit exists and is zero:
$$
\lim_{k\to\infty}\int f(kx)g(x)\,dm=0.
$$
(thanks to Norbert and to Nick Strehlke for the ideas to shorten the proof)
