Limit of Upper Darboux Sum I have a problem with a proof that I think should be trivial, but I am being unable to solve it. 
I have to show that if $f$ is bounded on $[a,b]$, and considering the "uniform partition $P_{n}$" in $[a,b]$, then
$  \int^b_a f = \lim_{n->\infty} U(P_{n},f)$
I know that this result, is a consequence of the "Dicing" Lemma, and that, then, this proof is straightforward. However, I am trying to proove it without using this lemma. 
I have two ideas,
1) I know that for each partition $\int^b_a f \leq U(P_{n},f) \leq U(P_{1},f)$, then the sequence given by $U(P_{n},f)$ is bounded and it is decreasing because of the "Refinement Lemma", so because of Monotone Convergence it should converge to its $lub=\int^b_a f$. However, I find that this idea is naive and I have an error somewhere.
2) The second idea is using somehow that $U(P_{n},f)=\sum_{i=0}^{n}M(f,[x_{i},x_{i+1}])(\frac{b-a}{n})=(\frac{b-a}{n})\sum_{i=0}^{n}M(f,[x_{i},x_{i+1}])$ and then taking the limit, but it is taking me nowhere.
Thank you very much!
 A: You need to understand that if $P_{n} = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ is a uniform partition of $[a, b]$ so that $x_{i} = a + ih, h = (b - a)/n$ then normally $P_{n + 1}$ is not a refinement of $P_{n}$. On the other $P_{2n}$ is a refinement of $P_{n}$. A partition $P'$ is a refinement of $P$ if $P \subseteq P'$.
You will perhaps now understand the importance of "Dicing Lemma". I will provide some remarks on it. Note that the usual definition of Riemann integral is as follows
Let $f$ be bounded on $[a, b]$. Then $f$ is said to be Riemann integrable on $[a, b]$ if there is a number $I$ such that for any given $\epsilon > 0$ it is possible to find a $\delta > 0$ which allows the following inequality $$\left|\sum_{i = 1}^{n}f(\xi_{i})(x_{i} - x_{i - 1}) - I\right| < \epsilon$$ to hold where $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ is any partition of $[a, b]$ with $||P|| = \max\{(x_{i} - x_{i - 1})\} < \delta$ and $\xi_{i}$ is any point in $[x_{i - 1}, x_{i}]$.
The behavior of the Riemann sum $\sum_{i = 1}^{n}f(\xi_{i})(x_{i} - x_{i - 1})$ is technically difficult to analyze when the norm $||P||$ of the partition is decreased. Hence an alternative and technically simple definition of Riemann integral is chosen which replaces the notion of "decreasing norm of a partition" with "refinement of a partition".
A refinement of a partition always leads to a decrease in the norm (or at best keep the norm unchanged), but the converse is not true so that decreasing the norm of a partition does not necessarily lead to a refinement.
The alternative definition goes as follows
Let $f$ be bounded on $[a, b]$. Then $f$ is said to be Riemann integrable on $[a, b]$ if there is a number $I$ such that for any given $\epsilon > 0$ it is possible to find a partition $P_{\epsilon}$ of $[a, b]$ which allows the following inequality $$\left|\sum_{i = 1}^{n}f(\xi_{i})(x_{i} - x_{i - 1}) - I\right| < \epsilon$$ to hold for any partition $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\} \supseteq P_{\epsilon}$ of $[a, b]$ and where $\xi_{i}$ is any point in interval $[x_{i - 1}, x_{i}]$.
Since we are switching to a technically simpler definition there might be some concern regarding loss of power of the concept being defined (or perhaps a risk that the concept itself has changed to something new). 
Dicing lemma says that such concerns are naive and both the definitions are equivalent. So you need to understand that "Dicing lemma" is not just some petty lemma used as a tool in the proof of some big significant theorem, but rather it is the "big significant theorem" itself. The result you are trying to prove is based on Dicing lemma and I doubt there is some fundamentally easier way to prove it.
