ProveUpper Riemann Integral is equal to Lower Riemann Integral, $U(f) = L(f)$ If $f$ is Riemann integrable, prove Upper Riemann Integral is equal to Lower Riemann Integral, $U(f) = L(f)$, where
$$U(f) = \inf \{ U(f, P) \mid P \text{ is a partition of } [a,b] \}$$
$$L(f) = \sup \{ L(f, P) \mid P \text{ is a partition of } [a,b] \}.$$
My professor asks us to prove this in parts.
(a) If $\varepsilon \gt 0 $, show that there is a partition of $P_1$ with width $W(P_1) \lt \delta$ and sample points for $P_1$ such that Riemann Sum $S_1 \gt U(f) -\frac{\varepsilon}{2}$ . (Hint: choose the partition based on denition of greatest lower bound, add points to make the width less than $\delta$)
points to make the width less
(b) show that there is a partition of $P_2$ with width $W(P_2) \lt \delta$ and sample points for $P_2$ such that Riemann Sum $S_2 \lt L(f) +\frac{\varepsilon}{2}$.
And another two parts..
I am having problems with proving both (a) and (b).
Edit: While it seems similar to this question, I am supposed to show that $U(f) = L(f)$ and in the addressed question, I don't think $U(f) = L(f)$ was proven.
 A: (I think that the inequalities in the hints are reversed of what they should be: for isntance, $U(f)$ is already less than all the upper sums)
Fix $\varepsilon>0$. Since $f$ is Riemann integrable, by your definition there exists a number $A$ andn $\delta>0$ such that $|S-A|<\varepsilon/4$ for every Riemann sum form a partition of width less than $\delta$. 
By definition of infimum there exists a partition $P_1$ such that $$\tag{1}U(f)+\frac\varepsilon4 > U(f,P_1).$$ By subdividing the partition, the upper sum either stays the same or decreases: if $m=\max\{f(t):\ x_1\leq t\leq x_2\}$ and $m_1=\max\{f(t):\ x_1\leq t\leq x_1'\}$, $m_2=\max\{f(t):\ x_1'\leq t\leq x_2\}$, then
$$
m_1(x_1'-x_1)+m_2(x_2-x_1')\leq m(x_1'-x_1)+m(x_2-x_1')=m(x_2-x_1).
$$
So we may assume, by subdividing $P_1$, that its width is less then $\delta$. This implies that 
$$\tag{2}
|U(f,P_1)-A|<\frac\varepsilon4.
$$
Similarly, by definition of supremum there exists a partition $P_2$ such that
$$\tag{3}
L(f)-\frac\varepsilon4<L(f,P_2).
$$
Again, by subdividing, we may assume that the width of $P_2$ is less than $\delta$ (note that subvididing the partition increases the lower sum, so the inequality $(3)$ is preserved. Again by definition of Riemman integrable, 
$$\tag{4}
|A-L(f,P_2)|<\frac\varepsilon4.
$$
Now
\begin{align}
|U(f)-L(f)|&\leq|U(f)-U(f,P_1)|+|U(f,P_1)-L(f,P_2)|+|L(f,P_2)-L(f)|\\ \ \\
&<\frac\varepsilon4+|U(f,P_1)-L(f,P_2)|+\varepsilon4\\ \ \\
&\leq\frac\varepsilon2+|U(f,P_1-A|+|A-L(f,P_2)|\\ \ \\
&<\frac\varepsilon2+\frac\varepsilon4+\frac\varepsilon4=\varepsilon.
\end{align}
As $\varepsilon$ was arbitrary, $U(f)=L(f)$. 
