Rana Measure Theory Theorem 1.1.4 Exercise This is a basic characterisation of Riemann integrability:
Given $f:[a,b] \rightarrow \mathbb{R}$ bounded the following are equivalent

*

*$f$ is Riemann integrable.

*For every $\epsilon > 0$ there exists a partition $P$ of $[a,b]$
such that $$U(P,f) - L(P,f) < \epsilon.$$

*There exists a unique real number $\alpha$ such that for every
partition $P$ of $[a,b]$ we have $$L(P,f) \leq \alpha \leq U(P,f).$$
Rana proves this by $1 \iff 2$ and $1 \iff 3$. After trying my hand at these proofs, I managed to get them all. But he has left one as an easy exercise that I am hoping to get feedback for:
$(2 \implies 1)$: Let $(P_n)$ be a sequence of partitions such that for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ with $$n \geq N \implies U(P_n,f) - L(P_n,f) < \epsilon.$$ Such a $(P_n)$ exists by assumption.
This gives a sequence of real numbers $(U(P_n,f))$ in $$\mathcal{U}:= \{U(P,f) : P \text{ is a partition }\}$$ and a sequence $(L(P_n,f))$ is a sequence in $$\mathcal{L}:=\{L(P,f) : P \text{ is a partition }\}.$$ Since both sequences have the same limit and $L(P_n,f) \leq U(P_n,f)$ for all $n \in \mathbb{N}$, we require $U(P_n,f)$ be decreasing while $L(P_n,f)$ must be increasing. So by the monotone convergence theorem we have
\begin{align*}
\lim_{n \rightarrow \infty} U(P_n,f) &= \inf \mathcal{U}\\
 \lim_{n \rightarrow \infty} L(P_n,f) &=\sup \mathcal{L}
\end{align*}
Once again, they limit to the same value allowing us to conclude $\inf \mathcal{U} = \sup \mathcal{L}$, this is the definition of $f$ being Riemann integrable.
 A: Let $\Pi$ the set of all possible partitions of $[a,b]$. Then let $\varepsilon >0$
$$\inf_{P\in\Pi}U(f,P)\leq U(f,P)  \ \ \ \ \forall P \in \Pi$$
$$\sup_{P\in\Pi}L(f,P)\geq L(f,P) \implies -L(f,P)\geq -\sup_{P\in\Pi}L(f,P) \ \ \ \ \forall P \in \Pi$$
By assumption, there exists $P_\varepsilon$ such that
$$\inf_{P\in\Pi}U(f,P)-\sup_{P\in\Pi}L(f,P)\leq U(f,P_\varepsilon)-L(f,P_\varepsilon)<\varepsilon$$
Since $\varepsilon$ is arbitrary*,
$$\inf_{P\in\Pi}U(f,P)=\sup_{P\in\Pi}L(f,P)$$
which means that $f$ is Riemann integrable.
(*) Let $|a-b|<\varepsilon$ for arbitrary $\varepsilon >0$. Suppose that $|a-b|=\varepsilon_0>0$. There exists a $\varepsilon$ such that $|a-b|<\varepsilon<\varepsilon_0$ which is a contradiction, thus $|a-b|=0\implies a=b$.

Feedback
Let $P_n$ be a sequence of partitions which satisfies
$$0 \leq U(f,P_n)-L(f,P_n)<\varepsilon_n \ \ \ \ \varepsilon_n \to 0$$
Define $a_n=U(f,P_n)-L(f,P_n)$. Since $a_n \to 0$ we have
$$\lim_{n \to \infty}(U(f,P_n)-L(f,P_n))=0$$
This alone does not guarantee they have limits. To make the argument surely work, you have to choose a sequence of refining partitions $P_n \subset P_{n+1}$. In this case, $U(f,P_n)$ is a decreasing sequence and $L(f,P_n)$ is an increasing sequence. In this case,
$$\inf_{n \in \mathbb{N}}U(f,P_n)-\sup_{n \in \mathbb{N}}L(f,P_n)\leq U(f,P_n)-L(f,P_n)<\varepsilon_n$$
