How to prove if $f$ is Darboux integrable then for all $\epsilon > 0$ then $U(f, P_{\epsilon}) - L(f, P_{\epsilon}) < {\epsilon}$ ??

Question

Background:

I am studying Real Analysis (never studied it before) from the book 'Real Analysis' by Jay Cummings. I am at chapter 8 (Integration) when I encounter theorem 8.14 which comes almost right after studying the definition of Darboux integral which author states is equivalent to Reimann integral (Pg. 268 under heading 'Type of Integrals'). which states the following:

Theorem in which I have doubt: Let $f: [a, b] \rightarrow \mathbb{R}$ be bounded. Then $f$ is integrable if and only if, for all $\epsilon > 0$ there exists a partition $P_\epsilon$ of $[a,b]$ where, $$U(f, P_\epsilon) - L(f, P_\epsilon) < \epsilon .$$

For reference I am also mentioning the definitions of integrable, partition and U and L here.

Partition: A partition of $[a,b]$ is a finite set $P = \{x_0, x_1, x_2, ..., x_n\}$ such that $a = x_0$, $b = x_n$ and $x_0 < x_1 < x_2 < .. < x_n$.

U (Upper Sum) and L (Lower Sum): Consider a function $f: [a,b] \rightarrow \mathbb{R}$ and consider a partition $P = \{x_0, x_1, ..., x_n\}$ of $[a,b]$. Define:

$$U(f, P) := \sum_{i=1}^{n} M_i \cdot (x_i - x_{i-1})$$ $$L(f, P) := \sum_{i=1}^{n} m_i \cdot (x_i - x_{i-1})$$

where $m_i$ := $\inf\{f(x) : x \in [x_{i-1}, x_i]\}$ and $M_i$ := $\sup\{f(x) : x \in [x_{i-1}, x_i]\}$ and

$$U(f) = \inf\{ U(f,P) : P ∈ \mathbb{P} \}$$ $$L(f) = \sup\{ L(f,P) : P ∈ \mathbb{P} \}$$

where $\mathbb{P}$ is the collection of all partitions of $[a,b]$.

Integrable: A bounded function $f: [a,b] \rightarrow \mathbb{R}$ is integrable if $L(f) = U(f)$.

Question:

In the proof of the above theorem I understood the forward direction but I am difficulty in understanding the intuition/validity behind one line (which I have made bold):

Let $\epsilon > 0$ and choose $P_{\epsilon}$ so that $U(f, P_\epsilon) - L(f, P_\epsilon) < \epsilon$. We know that $L(f, P_{\epsilon}) \leq L(f)$ and $U(f, P_{\epsilon}) \geq U(f)$. Which implies:

$$ U(f) - L(f) \leq U(f, P_{\epsilon}) - L(f, P_{\epsilon}) < {\epsilon} $$

But since ${\epsilon}$ was arbritary, this means that $U(f) - L(f) = 0$ and hence $L(f) = U(f)$, meaning that $f$ is integrable.

How did arbitrary $\epsilon$ implied that $U(f) - L(f)$ is equal to $0$??

I was thinking that if we think of it in terms of sequences of the form $U(f, P_{\epsilon_n}) - L(f, P_{\epsilon_n})$ then as $\epsilon_n$ is decreasing this sequence will converge to 0 (since it can be in any $\epsilon$ range of 0). Ok, lets say the point to which it converges is $U(f,P_{\epsilon_{\infty}}) - L(f,P_{\epsilon_{\infty}}) = 0$. Now:

1. Why should we even think that $P_{\epsilon_{\infty}}$ exists and is a valid partition?
2. Why would $U(f,P_{\epsilon_{\infty}}) - L(f,P_{\epsilon_{\infty}})$ be equal to $U(f) - L(f)$ ?

I do have some idea about this one maybe its because if $P_{\epsilon_{\infty}}$ does actually exist, then because $U(f) - L(f) \geq 0$ (Lemma 8.9 in the book says this is true for bounded functions) and through better partitions they will only end coming closer to each other i.e. $U(f)-L(f)$ will only decrease (I dont know how to prove this) and so $U(f)-L(f) = U(f,P_{\epsilon_{\infty}}) - L(f,P_{\epsilon_{\infty}}) = 0$ since it cant get better than 0.

Answer 1

You can observe the following: if $a$ be a real number such $0\le a<\epsilon$, for every $\epsilon$ then $a=0$. Let's go by contradiction,assume that the result is not true then $a\ne 0$, i.e $a>0$. If you take $\epsilon_{0}=\frac{1}{2} a$ then $\epsilon_{0}>0$ and $\epsilon_{0}<a$. In other hand by the hypothesis $0<a<\epsilon_{0}$. But this two are not true together so the assumption is wrong. Hence, the result is true.

In your case, consider $a=U(f)-L(f).$ So, if $U(f)-L(f)<\epsilon$ for every $\epsilon>0$ then $U(f)=L(f)$.