Please I have a slight confusion with the notation used by Rudin in the following proof.
11.33 Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and
$$ \int_a^b f \ dx = \mathscr{R} \int_a^b f \ dx $$
Where $\mathscr{R} \int$ denotes the Riemann integral, while $\int$ denotes the Lebesgue integral.Proof Suppose $f$ is bounded. Then there exists a sequence $\{P_k\}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and
$$ \lim_{k\to\infty} L(P_k,f) = \mathscr{R}\underline{\int_a^b} f \ dx, \quad \lim_{k\to\infty} U(P_k,f) = \mathscr{R}\overline{\int_a^b} f \ dx.
$$
Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k=\{a=x_0<x_1<\dots<x_n=b\},$ these are defined as,
$$ L(P_k,f) = \sum_{i=1}^n (x_i-x_{i-1})m_i, \quad U(P_k,f) = \sum_{i=1}^n (x_i-x_{i-1})M_i,$$
where $M_i = \sup_{x\in[x_{i-1},x_i]} f(x)$ and $m_i = \inf_{x\in[x_{i-1},x_i]} f(x).$We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x \in(x_{i-1},x_i],$ $1\leq i \leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $x\in [a,b],$
$$ L(P_k,f) = \int_a^b L_k \ dx, \quad U(P_k,f) = \int_a^b U_k \ dx, $$
and $$L_1(x) \leq L_2(x) \leq \dots \leq f(x) \leq \dots \leq U_2(x) \leq U_1(x). $$
There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x \in [a,b],$
$$ L(x) \leq f(x) \leq U(x), $$
and by the monotone convergence theorem,
$$
\int_a^b L(x) \ dx = \mathscr{R} \underline{\int_a^b} f \ dx, \quad \int_a^b U(x) \ dx = \mathscr{R} \overline{\int_a^b} f \ dx.
$$
Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) \leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.
I do understand the idea of the proof but the definition of the partition $P_k=\{a=x_0<x_1<\dots<x_n=b\}$. Shouldn't the end point depend on $k$ and not $n$?
Edit: By the explanation of @Ben W, how then is $$L_1(x) \leq L_2(x) \leq \dots \leq f(x) \leq \dots \leq U_2(x) \leq U_1(x). $$ obtained.
Best Answer
For each $k$ there is $n(k)$ such that $$P_k=\{x_0<x_1<\cdots<x_{n(k)}\}$$ However, it is more convenient just to write $n$ instead of $n(k)$.