Let $f:[a,b]\to\mathbb{R}$ be continuous. Consider, for all $n\in\mathbb{N}$, the homogeneous partition $P_n$ of order $n$. Prove that $\lim_{n\to\infty}L(f, P_n) = \int_a^b f=\lim_{n\to\infty}U(f, P_n)$.
So what I have tried is that since $f$ is continuous, then it's integrable. Meaning,
$$L(f, P_n)\leq sup(\{L(f, P)\mid P\in\mathbb{P}_{[a,b]}\}) = \int_a^b f= inf(\{U(f, P)\mid P\in\mathbb{P}_{[a,b]}\})\leq U(f, P_n).$$
Thus, now we must show that $\lim_{n\to\infty}U(f, P_n) = \lim_{n\to\infty}L(f, P_n)$. Indeed,
\begin{align*}
\lim_{n\to\infty}L(f, P_n)
&= \lim_{n\to\infty}\sum_{k=1}^n m_k(f)\lambda([t_{k-1}, t_k]) \\
&= \lim_{n\to\infty}\sum_{k=1}^n \left(m_k(f)\cdot\frac{b-a}{n}\right) \\
&= (b-a)\sum_{k=1}^n \left(\lim_{n\to\infty}\frac{m_k(f)}{n}\right)
\end{align*}
where $m_k(f) = inf\{f(x)\mid x\in[t_{k-1}, t_k]\}$ and $M_k(f) = sup\{f(x)\mid x\in[t_{k-1}, t_k]\}$. But from here, how do we get to the upper sums? I.e., how does
$$\lim_{n\to\infty}m_k(f) = \lim_{n\to\infty}M_k(f)?$$
Best Answer
Consider an arbitrary interval given by your partition, $I_k = [t_{k-1},t_k]$. Now, let $\frac{b-a}{n} = \Delta(x)$. Now choose your favorite $\epsilon > 0$, and note by uniform continuity of $f$ on $I_k$ that there is some $\delta > 0$ such that $|x - y| < \delta $ implies $|f(x) -f(y)| < \epsilon$.
Now, since $n \rightarrow \infty$, there is some rank $N$ such that $m \geq N \rightarrow \Delta(x) = \frac{b-a}{m} < \delta$, since $\delta$ is fixed. But then, consider $I_k$: note that for all $x,y \in I_k, |x - y| \leq \Delta(x) < \delta$. However, this implies that for all $x,y \in I_k, |f(x) - f(y)| < \epsilon$. Now, from what I initially commented, $m_k(x) = f(x_1^k), M_k(x) = f(x_2^k)$ for some $x_1^k, x_2^k \in I_k$. Thus, specialising our finding, $|f(x_1^k) - f(x_2^k)| < \epsilon$. This gives us our claim.