# Sequence of divisions such that its upper and lower function limits are equal

Let $A\ne \emptyset$ be a Jordan measurable subset of $\mathbb{R}^n$ and $f: A \to \mathbb{R}$ be a bounded function, such that $f$ is integrable. Given that for every $\varepsilon >0$, there exists a division $\Delta$ of $A$ such that $U( f , \Delta ) - L( f, \Delta ) < \varepsilon$, I want to prove that there exists a sequence $( \Delta_k )_{k=1}^{\infty}$ of divisions of $A$ such that $\lim\limits_{k\to\infty} U( f , \Delta_k ) - L( f, \Delta_k ) = 0$.

My approach is as follows:

Let $(\varepsilon_k)_{k=1}^\infty$ be a sequence of positive numbers, such that $(\varepsilon_k)_{k=1}^\infty$ converges to $0$. Then for every $k\ge 1$ there exists a division $\Delta_k$ such that $U(f, \Delta_k)-L(f,\Delta_k)<\varepsilon_k$. Then for every $\varepsilon'>0$, $\exists k_0\in\mathbb{N}$ such that $k\ge k_0$ implies $\varepsilon-\varepsilon_k<\varepsilon'$. This implies that $U(f, \Delta_k)-L(f,\Delta_k)<\varepsilon' + \varepsilon$. Since $\varepsilon$ is arbitrary, this implies that $\lim\limits_{k\to\infty} U( f , \Delta_k ) - L( f, \Delta_k ) = 0$.

I'd appreciate if you could please comment on my proof. Do you think it actually needs improvement?

If $\epsilon_k \to 0$ and $U(f,\Delta_k) - L(f,\Delta_k) < \epsilon_k$ then you can simply apply the squeeze theorem since $U(f,\Delta_k) - L(f,\Delta_k) \geqslant 0$ always.
Cleaning up your argument, for every $\epsilon > 0$ there exists $k_0$ such that for $k \geqslant k_0$ we have $\epsilon_k < \epsilon$ and
$$|U(f,\Delta_k) - L(f,\Delta_k)| = U(f,\Delta_k) - L(f,\Delta_k) < \epsilon_k < \epsilon$$
• Thanks. Do you think my proof might still hold some water? I'm trying to find out if there could possibly be some logical error. Here's the definition that I have: "A set $A \subseteq \mathbb{R}^n$ is Jordan measurable when it satisfies the following two conditions: (i) $A$ is a bounded set; (ii) the boundary bd$(A)$ is a zero-content set." – sequence Jul 11 '17 at 4:20