The infimum over all partitions is the same as the infimum over all partitions including a fixed partition $Q$

42 Views Asked by At

Take an interval $[a,b]$ on the real line and a bounded function $f:[a,b] \to \mathbb R$. For a partition $P =\{ a = t_0, t_1, \dots, t_n = b\}$ of said interval we define the Upper Darboux Sum of $f$ on the partition $P$ as

$$U(f,P) := \sum_{i=1}^n \sup_{t \in [t_{i-1}, t_i]} f(t) \cdot (t_i - t_{i-1})$$

We know that Upper Darboux Sums are decreasing with respect to refinements, so $P \subseteq P' \Rightarrow U(f,P) \ge U(f,P')$

I want to prove that for a fixed partition $Q$ of $[a,b]$, it's true that

$$\inf_P \;U(f,P) = \inf_P \;U(f, P \cup Q) = \inf_{P \supseteq Q} U(f,P)$$

The result is pretty "intuitive" but I don't know how to prove it formally

From the inequality $U(f,P) \ge U(f, P \cup Q)$ we immediately get the inequality

$$\inf_P \;U(f,P) \ge \inf_P \;U(f, P \cup Q) = \inf_{P \supseteq Q} U(f,P)$$

I'm struggling to either prove the other inequality or prove they must be equal

1

There are 1 best solutions below

1
On BEST ANSWER

The set of all partitions finer than $Q$ is a subset of all partitions. Taking the infimum over only some partitions is $\ge$ than taking the infimum over possibly more partitions