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 $\mathbb{R}^n$.
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 $\mathbb{R}^n$.
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)