Prove Upper Darboux Integral

[Yoni V]

Homework Statement

Let $f$ be a bounded function on $[0,1]$. Let $P_n$ be a partition of $[0,1]$ such that $P_n = (0,\frac{1}{n},\frac{2}{n},...,1)$. Finally, we define $\alpha=\inf\{U(f,P_n):n\geq1\}$, where $U(f,P)$ is the upper Darboux sum of $f$ with partition $P$.

Show that $\alpha = U_f$, with $U_f$ being the upper Darboux integral of $f$, i.e. $U_f = inf\{U(f,P):P\;is\;a\;partition\;of\;[0,1]\}$.

Homework Equations

The Attempt at a Solution

We are also given the hint, which we showed in class, that for every partition $P$,
$$U(f,P\cup \{s \})-(M_f-m_f) \delta (P) \geq U(f,P)$$ where $s \in [0,1]$, $M_f=max(f)$ on [0,1] (and correspondingly $m_f$) and $\delta(P)$ is the norm of P.

I really can't make the first step. I was thinking of using a theorem about the upper integral $\bar I=\lim_{\delta(p) \rightarrow0}U(f,P)$, but I can't get it to take me anywhere...

Thanks

 

[mighty2000]

for your post! This is an interesting problem and it requires some careful analysis to prove the statement. Let's start by defining some key terms and concepts.

First, let's define what a partition is. A partition of an interval [a, b] is a finite sequence of numbers a = x0 < x1 < ... < xn = b. Each subinterval [xi-1, xi] is called a subinterval of the partition. In this problem, the partition Pn is defined as (0, 1/n, 2/n, ..., 1), which means that the interval [0,1] is divided into n subintervals of equal length. This is important because it allows us to control the size of the subintervals and make them smaller as n gets larger.

Next, let's define what an upper Darboux sum is. The upper Darboux sum of a bounded function f on an interval [a, b] with respect to a partition P is defined as U(f, P) = ∑(xi - xi-1)M(xi), where M(xi) is the maximum value of f on the subinterval [xi-1, xi]. This is essentially the area of rectangles that lie above the graph of f on each subinterval.

Now, let's define the upper Darboux integral. The upper Darboux integral of a bounded function f on an interval [a, b] is defined as Uf = inf{U(f, P): P is a partition of [a, b]}. This is essentially the smallest possible upper Darboux sum for all possible partitions of the interval [a, b].

Now, let's look at the hint given in the problem. It tells us that for any partition P, we have U(f, P∪{s}) - (Mf - mf)δ(P) ≥ U(f, P), where s is a point in [0, 1], Mf and mf are the maximum and minimum values of f on [0, 1], and δ(P) is the norm of P. This is a useful inequality because it tells us that for any partition, the upper Darboux sum will be at least as large as the upper Darboux sum for a partition with an extra point added, minus the difference between the maximum and minimum values of f on [0, 1] times the norm of the partition.

Now, let