Show the lower Riemann integral is the limit of lower sums of a sequence of refined partitions I just finished the first seven chapters of Baby Rudin, and start working on Measure, Integration, and Real Analysis by Sheldon Axler. The following is Exercise 7 of Section 1A.

Suppose $f\colon [a,b]\to \mathbb{R}$ is a bounded function. For $n\in \mathbb{Z}^+$, let $P_n$ denote the partition that divides $[a,b]$ into $2^n$ intervals of equal size. Prove that $$L(f,[a,b])=\lim_{n\to\infty}L(f,P_n,[a,b])\quad \text{and}\quad U(f,[a,b])=\lim_{n\to\infty}U(f,P_n,[a,b]).$$

For any finite partition $P=\{a=x_0<\cdots<x_n=b\}$ of $[a,b]$, the lower and the upper Riemann sums are defined as \begin{align*}
L(f,P,[a,b])&:= \sum_{i=1}^n\left(\inf_{t\in[x_{i-1},x_i]}f(t)\right)(x_i-x_{i-1}),\\
U(f,P,[a,b])&:=\sum_{i=1}^n\left(\sup_{t\in[x_{i-1},x_i]}f(t)\right)(x_i-x_{i-1}).
\end{align*}
The lower Riemann integral is $L(f,[a,b]):= \sup L(f,P,[a,b])$, where the supremum is taken over all partitions. Similarly, $U(f,[a,b]):=\inf U(f,P,[a,b])$.
My attempts: That $P_n\subset P_{n+1}$ for any $n$ implies $L(f,P_n,[a,b])\le L(f,P_{n+1},[a,b])$ for any $n$, which implies $\lim_{n\to\infty}L(f,P_n,[a,b]) = \sup_n L(f,P_n,[a,b])\le L(f,[a,b])$. To show the converse direction, it suffices to show the following proposition:

Given any partition $P$, there exists $n$ such that $L(f,P,[a,b])\le L(f,P_n,[a,b])$.

Let $P=\{y_0<\cdots<y_m\}$ be given. Let $n$ be sufficiently large and take $P_n=\{x_0<\cdots <x_{2^n}\}$. Let $I\subseteq \{1,...,2^n\}$ be the indices for which $[x_{i-1},x_i]$ is contained in some $[y_{j-1},y_j]$. Then $\inf_{t\in[x_{i-1},x_i]}f(t)\ge \inf_{t\in[y_{j_i-1},y_{j_i}]}f(t)$ for all $i\in I$. The $\ge$ is what I want. However, I failed to find a lower bound for $i\in I^c$. The only thing I know is that $i\in I^c$ only if $y_j\in(x_{i-1},x_i)$ for some $j$, so that $|I^c|\le m-1$. But this is a $\le$, not a $\ge$.
I appreciate any help.
 A: I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
This exercise is Exercise 7 in Exercises 1A on p.8.
If $f$ is a constant function, it is easy to prove the result.
So, we assume that $f$ is not constant.
Let $\epsilon$ be an arbitrary positive real number.
Let $P=x_0,x_1,\dots,x_n$ be a partition of $[a,b]$ such that $U(f,[a,b])+\epsilon>U(f,P,[a,b])$.
Let $\delta$ be a positive real number such that $\min\{x_1-x_0,x_2-x_1,\dots,x_n-x_{n-1}\}>\delta$.
Let $N$ be a positive integer such that $\min(\delta,\frac{\epsilon}{(n+1)(\sup_{[a,b]} f-\inf_{[a,b]} f)})>\frac{b-a}{2^N}$.
Let $\delta':=\frac{b-a}{2^N}$.
Let $P'=x_0,x_0+\delta',x_0+2\delta',\dots,x_0+(2^N-1)\delta',x_n$.
For any $i\in\{1,\dots,2^N\}$, $[x_0+(i-1)\delta',x_0+i\delta']$ contains no element of $\{x_0,\dots,x_n\}$ or only one element of $\{x_0,\dots,x_n\}$.
Let $m_j\in\{0,\dots,2^N-1\}$ be an integer such that $x_j\in [x_0+m_j\delta',x_0+(m_j+1)\delta']$ for each $j\in\{0,\dots,n\}$.
Let $P^{''}$ be the partition of $[a,b]$ obtained by merging the lists that define $P$ and $P'$.
$$U(f,P',[a,b])-U(f,P^{''},[a,b])=\\\sum_{j=0}^{n} \delta'\times\sup_{[x_0+m_j\delta',x_0+(m_j+1)\delta']} f-\\\left(\sum_{j=0}^{n} (x_j-(x_0+m_j\delta'))\times\sup_{[x_0+m_j\delta',x_j]} f+\\\sum_{j=0}^{n} (x_0+(m_j+1)\delta'-x_j)\times\sup_{[x_j,x_0+(m_j+1)\delta']} f\right)\\<\sum_{j=0}^{n}\delta'\times\sup_{[a,b]} f-\\\left(\sum_{j=0}^{n} (x_j-(x_0+m_j\delta'))\times\inf_{[a,b]} f+\\\sum_{j=0}^{n} (x_0+(m_j+1)\delta'-x_j)\times\inf_{[a,b]} f\right)\\=\sum_{j=0}^{n}\delta'\times (\sup_{[a,b]} f-\inf_{[a,b]} f)\\=(n+1)\times (\sup_{[a,b]} f-\inf_{[a,b]} f)\times\delta'<\epsilon.$$
So, $U(f,[a,b])+2\epsilon>U(f,P,[a,b])+\epsilon\geq U(f,P^{''},[a,b])+\epsilon>U(f,P',[a,b])$.
So, $U(f,[a,b])=\lim_{n\to\infty} U(f,P_n,[a,b])$.
Similarly, $L(f,[a,b])=\lim_{n\to\infty} L(f,P_n,[a,b])$.
