While reading this topic, I had a stuck. How to show these two definitions of the Riemann integral are equivalent?
In the answer, they said that "analoguosly, $L(P, f)>I-\epsilon$", but I can't see why it's analoguous. I've tried to put $L(P, f)$ into two sums $S_{1}$, $S_{2}$ like $U(P, f)$ but then I can't compare $S_{1}$ with $L(P_{\epsilon}, f)$. Can anybody help me?
Sorry if my question is silly.
You are right to separate the sum into $S_1$ and $S_2$. The infimum in each term appearing in $S_1$ increases. Regarding the length of the subintervals, although the subintervals of the partition $P$ appearing in $S_1$ have smaller length than the corresponding subintervals of the partition $P_{\epsilon }$, the total length of all $P$-subintervals that are strictly inside a $P_{\epsilon }$-subinterval is almost equal (up to a $2\delta $ difference) to the length of the $P_{\epsilon }$-subinterval.
Here is a precise argument to get an inequality with $2\epsilon $ (but it's possible to get just $\epsilon $, with some extra effort):
If $P_{\epsilon }=\{ x_1,\ldots ,x_N\} $, then denote by $x_{k,1},\ldots ,x_{k,N_k}$ the points of the partition $P$ that lie between $x_{k}$ and $x_{k+1}$. Then $$ S_1=\underset{k}{\sum }\underset{i}{\sum }m_{k,i}\Delta x_{k,i}\geq \underset{k}{\sum }\underset{i}{\sum }m_{k}\Delta x_{k,i}=\underset{k}{\sum }m_{k}\underset{i}{\sum }\Delta x_{k,i}\geq \underset{k}{\sum }m_{k}(\Delta x_k-2\delta )=\underset{k}{\sum }m_{k}\Delta x_k-2\delta \underset{k}{\sum }m_{k}\geq \underset{k}{\sum }m_{k}\Delta x_k-2NM\delta =L(f,P_{\epsilon })-\epsilon $$ Since $S_2\geq -\frac{\epsilon }{2}$ and $L(f,P_{\epsilon })>I-\frac{\epsilon }{2}$, it follows that $S_1+S_2>I-2\epsilon $.
