Real Analysis, Integration 
Let $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ be a continuous function such that $\lim_{|x|\rightarrow \infty} f(x)=0$. Prove that $\lim_{k\rightarrow \infty}\int_{[0,1]^{n}}f(kx)dx=0. $

I don't know how to proceed, anyone have any idea? Thanks!!
My idea: If $\lim_{|x|\rightarrow \infty} f(x)=0$, then $\lim_{|kx|\rightarrow \infty} f(kx)=0$, then for all $\epsilon>0$, $|kx|>A\ , (A>0)$, then $|f(kx)|<\epsilon $. Now, $\Big|\int_{[0,1]^{n}}f(kx)dx\Big|\leq \int_{[0,1]^{n}}|f(kx)|dx \leq \epsilon \ vol([0,1]^n)=\epsilon .$

Hint: Since $f$ is continuous and $\lim_{\lvert x\rvert\to\infty}f(x)=0$, we have $f$ is bounded and uniformly continuous.  Now what is the obvious thing we can do to $\int_{[0,1]^n} f(kx)\,\mathrm{d}x$?
