Problem involving lower and upper sums in intergration 
If $f$ is integrable on $[a,b]$, prove that for any partition $P$, $LS(f,P)\leq \int_{a}^{b}f \leq   US(f,P)$.

My attempt: Since $f$ is integrable on $[a.b]$, by definition there is some number $I\in \mathbb{R}$ so that for all $\epsilon>0$ there is a $\delta>0$ so that for any partition $P$ with $\left \| P \right \|< \delta $ and any $S$ sample points, $|RS(f,P,S) - I|< \epsilon$.
So $RS(f,P,S) - \epsilon \leq I \leq RS(f,P,S) + \epsilon \Rightarrow  LS(f,P) - \epsilon\leq RS(f,P,S) - \epsilon \leq I \leq RS(f,P,S) + \epsilon \leq US(f,P) + \epsilon$.
Then where do I go from here? 
 A: Standard approach is to recognize -
$1) L(f,P)\le U(f,P)$(which is clear from the definition)
$2)L(f,P)\ge L(f,P_1) \text{ and } U(f,P_1)\ge U(f,P)$ if $P$ is a refinement of $P_1$.
This is due to the fact that if an interval $[x_i,x_{i+1}]$ is divided into two sub-intervals then one of them may not contain the point in $[a,b]$ where f is maximum and therefore refining intervals can only lessen the upper sum if it is changed at all and similar reasoning shows refining intervals can not decrease the lower sum.
$3)$If $P_1$ and $P_2$ are arbitary partitions of the same interval, we can choose a partition $P$ that is a refinement for both $P_1$
and $P_2$.
So if $P_1,P_2,P$ are partitions as in $(3)$ we have from $(2)$ $L(f,P)\ge L(f,P_1) \text{ and } U(f,P)\le U(f,P_2)$ and from $(1)$ we have $ L(f,P)\le U(f,P)$. Combining these we have $L(f,P_1)\le L(f,P)\le U(f,P)\le U(f,P_2)\Rightarrow L(f,P_1)\le U(f,P_2)$. As $P_1$ and $P_2$ are arbitary partitions the last inequality implies sup{($L(f,P)$)}$\le$ inf{$U(f,P)$}. Therfore, for an integrable function, from the definition of R-integrability we have
$L(f,P)\le$ sup{($L(f,P)$)}=$\int_a^bf=$inf{$U(f,P)$}$\le U(f,P)$. 
