Proving a bounded function is integrable I have a homework question which is:

If $f(x)$ is a bounded function in $[0,1]$ and $\sup
 (f[\frac {1}{n},1])-\inf (f [\frac {1}{n},1])<\frac {1}{n}$ for
  every natural $n>0$. Prove that $f(x)$ is Integrable in $[0,1]$.

I have a feeling that this can be proven by showing somehow that $\inf(S_n-s_n)=0$ where $S_n$ and $s_n$ are the upper and lower limits of the split $P_n$.
However I have not managed to prove this, perhaps I am attacking this problem from the wrong direction. 
Can someone help me out? 
Thanks :) 
 A: If you are familiar with the terminology of baby rudin (where the darboux integral is set up), the following reasoning might work. For every $n$ let $P_n=\{0,1/n,1\}$. Furthermore let $M_1=sup_{x\in[1/n,1]} f(x)$
and $M_0=sup_{x\in[0,1/n]} f(x)$
and similarly let  $m_1=inf_{x\in[1/n,1]} f(x)$ and  $m_0=inf_{x\in[0,1/n]} f(x)$
Now in the terminology of baby rudin we have:
$U(P_n,f)=(1/n)M_0+(1-1/n)M_1$ and $L(P_n,f)=(1/n)m_0+(1-1/n)m_1$
Let $\epsilon > 0$ and choose $n>2(M_0-m_0)/\epsilon$ and $(n-1)/n^2<\epsilon/2 ,$ then 
$U(P_n,f)-L(P_n,f)=(M_0-m_0)(1/n)+(M_1-m_1)(1-1/n)<\epsilon$ 
A: Fix $\varepsilon>0$ and $n$ such that $\max\left\{1,2\cdot\sup_{[0,1]}f\right\}n^{-1}\leq\varepsilon$. Consider the subdivision $0<n^{-1}<1$ and put $s(x)=\begin{cases}\inf_{0\leq t\leq n^{-1}}f(t)&\mbox{ if }0\leq x\leq n^{-1}\\\
\inf f\left([n^{-1},1]\right)&\mbox{ if }n^{-1}< x\leq 1,\end{cases}$ and $S(x)=\begin{cases}\sup_{0\leq t\leq n^{-1}}f(t)&\mbox{ if }0\leq x\leq n^{-1}\\\
\sup f\left([n^{-1},1]\right)&\mbox{ if }n^{-1}< x\leq 1,\end{cases}$
