I'm a bit stuck with this problem and was wondering if someone could help me out.
Define $f\colon [0,1]\to \mathbb{R} $ by $f(x)=x.$
(a) Prove that for any partition $P$ of $[0,1]$ we have $U(f,P) \gt \frac{1}{2}.$
(b) Prove that for each $\epsilon \gt 0$ there is a partition $P_{\epsilon}$ of $ [0,1]$ such that $U(f,P_{\epsilon}) \le \frac{1}{2}+\epsilon.$
(a)
Let $n \in \mathbb{N}$. Let $P_{n}=\{\frac{0}{n}\frac{1}{n},\frac{2}{n},...,\frac{n}{n}\}$.
Consider $U(f,P_{n})=\sum_{i=1}^{n}M_{i}(x_{i}-x_{i-1})=\sum_{i=1}^{n}\frac{i}{n}\frac{1}{n}=\frac{1}{2}+\frac{1}{2n} \gt \frac{1}{2}$.
And then for (b) I was thinking that since the collection of all upper Darboux sums is decreasing, by the Refinement Lemma, and bounded, I can use the Monotone Convergence Theorem to conclude that $U(f)$, i.e., the $inf$ of all upper Darboux sums is $\frac{1}{2}$. And so the rest follows.
But I know there is something really wrong with my solution. I guess starting with the fact that I haven't considered all possible partitions ... but only those with subintervals of equal length. How would you solve this problem for any partition $P$ of $[0,1]?$
Edit: I'm thinking that maybe such partitions $P_{n}$ are subsets of all possible partitions of [0,1] so in considering $P_{n}$ I am also considering all possible partitions?
a) is not correct, you cannot fix a partition that $P_{n}=\{0,1/n,...,n/n\}$. Rather, for any partition $P=\{0=x_{0},...,x_{n}=1\}$, $M_{i}=\sup_{x\in[x_{i-1},x_{i}]}f(x)=\sup_{x\in[x_{i-1},x_{i}]}x=x_{i}$, so \begin{align*} U(f,P)&=\sum_{i=1}^{n}M_{i}(x_{i}-x_{i-1})\\ &=\sum_{i=1}^{n}x_{i}(x_{i}-x_{i-1})\\ &>\sum_{i=1}^{n}\dfrac{1}{2}(x_{i}+x_{i-1})(x_{i}-x_{i-1})\\ &=\dfrac{1}{2}\sum_{i=1}^{n}(x_{i}^{2}-x_{i-1}^{2})\\ &=\dfrac{1}{2}(x_{n}^{2}-x_{0}^{2})\\ &=\dfrac{1}{2}. \end{align*}
b) Given $\epsilon>0$, choose a positive integer $n$ such that $1/n<\epsilon$, then for the partition that $P=\{0,1/n,...,n/n\}$, we have \begin{align*} U(f,P)&=\sum_{i=1}^{n}\dfrac{1}{n}\cdot\dfrac{i}{n}\\ &=\dfrac{1}{n^{2}}\cdot\dfrac{n(n+1)}{2}\\ &=\dfrac{1}{2}+\dfrac{1}{2n}\\ &<\dfrac{1}{2}+\epsilon. \end{align*}