I was wonderind why the function $f(x)=x, x \in [0,1]$ is integrable in $[0,1]$,although $U(f,P) \neq L(f,P)$
$P$ a partition of $[0,1]$,let $P=\{ x_0=0,x_1=1\}$
$U(f,P)=(1-0) \sup f([0,1])=1$
$L(f,P=(1-0) \inf f([0,1])=0$
$\overline \int_0^1 f=1 \neq \underline \int_0^1 f=0$
Your logic is incorrect. The upper Darboux integral in English is the result of $U(f,P)$ where $P$ is the partition that results in a smallest value, and the lower Darboux integral is $L(f,P)$ where $P$ produces the highest value. So, you give:
$$\overline \int_0^1 f=1 \neq \underline \int_0^1 f=0$$
This is not true, to show you a counterexample: Let $P = \{[0,\frac{1}{2}], [\frac{1}{2},0]\}$ Notice that the $U(f,P) = 1/4+1/2 < 1$ and $L(f,P) = 0+1/4 > 0$.
Note that
$$ U(f,P) = \sum_{i=1}^n (x_i - x_{i-1}) \sup_{x \in [x_{i - 1}, x_i]} f(x) \\ L(f,P) = \sum_{i=1}^n (x_i - x_{i-1}) \inf_{x \in [x_{i - 1}, x_i]} f(x) \\ \overline{\int_a^b} f(x) \, \mathrm{d} x = U(f,[a,b]) = \inf \{U(f,P) \mid P \text{ a partition of } [a,b]\} \\ \underline{\int_a^b} f(x) \, \mathrm{d} x = U(f,[a,b]) = \sup \{L(f,P) \mid P \text{ a partition of } [a,b]\} $$
So indeed for your chosen $P$ you have the correct $U(f,P),L(f,P)$ however you don't have the correct $\overline{\int_0^1} f, \underline{\int_0^1} f$ since you did not take the inf and sup respectively.
EDIT: Let's talk about how we can get some value for $U(f,[a,b])$.
First we should talk about refinements of a partition. Given a partition $P$ a refinement of this partition $P'$ if every endpoint of $P$ is also in $P'$ (that is you can only add more endpoints, not remove any). It is then known that $$ U(f,P) \ge U(f,P') \\ L(f,P) \le L(f,P') $$ so that if we refine a partition we get a smaller upper sum and a larger lower sum.
Now let's start refining your partition of $x_0 = 0,x_1 = 1$ by inserting a $\frac{1}{2}$ so that $P = \{ 0, 1 / 2, 1 \}$. We can continue to subdivide these subintervals to form partitions $P_n = \left\{ \frac{k}{n} \mid k = 0,1,\ldots n\right\}$ now let's look at $U(f,P_n)$ for a general $n$:
$$ \begin{eqnarray} U(f,P_n) & = & \sum_{i=1}^n \left( \frac{i}{n} - \frac{i-1}{n} \right) \frac{i}{n} \\ & = & \sum_{i = 1}^n \frac{1}{n} \frac{i}{n} \\ & = & \frac{1}{n^2} \sum_{i=1}^n i \\ & = & \frac{1}{n^2} \frac{n(n-1)}{2} \\ & = & \frac{n^2 - n}{2n^2} \end{eqnarray} $$ Now in theory since each refinement produces a smaller upper sum if we take these refinements to infinity ($n \to \infty$) we may obtain the $\inf$. So doing that we get $$ \lim_{n \to \infty} U(f,P_n) = \lim_{n \to \infty} \frac{n^2 - n}{2n^2} = \frac{1}{2} $$ Now all that's left is proving indeed that this is the $\inf$. So, you need to show that this is a lower bound and then as usual that any other lower bound is $\le \frac{1}{2}$ (Hint: suppose that there is some number $\frac{1}{2} + \epsilon$ that is also a lower bound but show that there is some partition $P$ so that $U(f,P) < \epsilon + \frac{1}{2}$).
In general I'm unsure if taking $n \to \infty$ of $U(f,P_n)$ will result in the $\inf$.
Consider that $$ U(f,P_n) \to L \text{ as } n \to \infty $$ Then surely we have that there is an $N$ so that $U(f,P_N) < \epsilon + L$ since we have for any $\epsilon > 0$ there is some $N$ so that $\lvert U(f,P_N) - L \rvert < \epsilon$. However I cannot see right now why $L$ is indeed a lower bound. Sure for any $P_n$ it's a lower bound but for any partition at all I'm unsure. I believe this is true but I cannot come up with a proof right now.
Hope this helps!
