Showing $\lim_{n \to \infty} L(f,P_n,[a,b]) = L(f,[a,b])$, where $P_n$ is partition of $[a,b]$ into $2^n$ subintervals Suppose $f : [a,b] \to \mathbb{R}$ is bounded. With a partition $P$ of the form $a = x_0, \dots ,x_n = b$ of $[a,b]$, the lower Riemann sum is $L(f,P,[a,b]) := \sum_{i=1}^{n} (x_i - x_{i-1}) \inf_{[x_{i-1},x_i]} f$. Then the lower Riemann integral is $L(f,[a,b]) := \sup_{P} L(f,P,[a,b])$; that is, the lower Riemann integral is the supremum over all the lower Riemann sums. Define the sequence $L(f,P_n,[a,b])$, where $P_n$ is the partition of $[a,b]$ obtained by splitting $[a,b]$ up into $2^n$ intervals of equal size. I want to prove that $\lim_{n \to \infty} L(f,P_n,[a,b]) = L(f,[a,b])$.
Here's my approach so far: Since the list defining partition $P_{n-1}$ is a sublist of the list defining the partition $P_n$, we have $L(f, P_{n-1}, [a,b]) \leq L(f,P_n, [a,b])$. That is, the sequence $L(f,P_n,[a,b])$ is monotone non-decreasing. Since it also has an upper bound of $L(f,[a,b])$, it follows from the monotone convergence theorem that $L(f,P_n,[a,b])$ is a convergent sequence, and it converges to the least upper bound of its terms. I need to prove that this limit is equal to $L(f,[a,b])$. This is where I am getting stuck.
Let the limit of $L(f,P_n,[a,b])$ be $l$. It is immediate that $l \leq L(f,[a,b])$, because the supremum of a subset is at most the supremum of the original set. So I need just to show that $L(f,[a,b]) \leq l$ to complete the proof. It is equivalent to show that for all $\epsilon >0$, we have $L(f,[a,b]) > l - \epsilon$. To this end, let $\epsilon > 0$. Since $L(f,[a,b])$ is the supremum of the lower Riemann sums, there exists a partition $P$ of $[a,b]$ such that $L(f,P,[a,b]) > L(f,[a,b]) - \frac{\epsilon}{2}$. If I can show that there exists $N \in \mathbb{N}$ such that $L(f,P_N, [a,b]) \geq L(f,[a,b])$ and $|L(f,P_N,[a,b]) - l| < \frac{\epsilon}{2}$, then I am done, since I will have as a consequence that $|l - L(f,[a,b])| < \epsilon$, by the triangle inequality.
Intuitively, I want to use the triangle inequality to show a connection between four 'things'. Firstly, I have the sequence $L(f,P_n,[a,b])$, which is increasing (or at least non-decreasing) to $L(f,[a,b])$ as $n$ gets big. I know I can approximate $L(f,[a,b])$, the second thing, with a margin $\epsilon$ of error, by $L(f,P,[a,b])$, the third thing, for some partition $P$. Then I just want to show that if I go far enough into the sequence $L(f,P_n,[a,b])$, the terms eventually get at least as big as $L(f,P,[a,b])$. Once they are at that threshold, the terms are within an $\epsilon$ margin of error to $L(f,P,[a,b])$, and using the triangle inequality to get an upper bound on the distance between $l$, the fourth thing, and $L(f,[a,b])$, I would be finished. But how do I do this? 
 A: The essential thing here is that $\|P_n\|\to 0$.
Use the standard inequality $$L(f,P)-L(f,Q) \le 2\,M\,p\, \|Q\|$$ where $M$ is an upper bound of $|f|$ and $p$ is the number of partition points of $P$.
A: Following up on your idea (which is good btw), I think you are only missing one small link in your chain of thought.
You are approximating $L(f, [a,b])$ by $L(f,P ,[a,b])$ and you'd like to approximate  $L(f,P, [a,b])$ by $L(f, P_n, [a,b])$. To do this, assume first that your partition $P$ looks like $P :  x_0=a < x_1 <...<x_k=b$ for some fixed k. Then note that if n is sufficiently large, the intervals in the partition $P_n$ are either contained in the intervals $[x_i, x_{i+1}]$ from P, or they overlap with two consecutive intervals of the form  $[x_i, x_{i+1}]$, $[x_{i+1}, x_{i+2}]$ from P. But(here is the key!) note that the overlaps are negligible(they can be made arbitrarily small) because f is bounded and because the lengths of these overlaps go to 0. Also note that if an interval from the $P_n$ partition is contained in an interval from $P$ its infimum is larger.
A: Let $P = \{ t_0,\dots,t_n \}$ be a partition of $[a,b]$, that is, $a = t_0 < t_1 < \dots < t_n = b$. Let $N$ be a sufficiently large integer; in particular, let $2^{-(N+1)} > d$, where $d = \max \{ t_i - t_{i-1} : 1 \leq i \leq n \}$. Let $P_N = \{ s_0,\dots,s_{2^N} \}$.
If for each $i$ we had that $t_i = s_{k}$ for some $k = k(i)$, then $P_N$ would be a refinement of $P$ and we could easily compare the the summations of $L(f,P,[a,b])$ and $L(f,P_N,[a,b])$ to conclude that $L(f,P_N,[a,b]) > L(f,P,[a,b])$.
The difficulty arises when $t_i \in (s_{k(i) - 1}, s_{k(i)})$ for some $k(i)$. In this case, what we can show is that $L(f,P,[a,b]) \leq L(f,P_N,[a,b]) + \mathcal{E}_N$, where $\mathcal{E}_N \to 0$ as $N \to \infty$. This will imply that $L(f,P,[a,b]) \leq l$, so if we had chosen the partition $P$ so that $L(f,P,[a,b]) > L(f,[a,b]) - \epsilon$, then we have $L(f,[a,b]) < l + \epsilon$ as desired.
The way to show that $L(f,P,[a,b]) \leq L(f,P_N,[a,b]) + \mathcal{E}_N$ is to compare the partition in $P_N$ lying between $t_i$ and $t_{i+1}$ and then dealing with the end points separately. The complete solution is given below.

Let $m_j = \inf \{ f(x) : s_{j-1} \leq x \leq s_{j} \}$ and $\tilde{m}_j = \inf \{ f(x) : t_{j-1} \leq x \leq t_j \}$. Let us start by comparing $$(t_i - t_{i-1}) \tilde{m}_i$$ with $$\sum_{k(i-1) \leq j \leq k(i)} (s_j - s_{j-1}) m_j.$$
  Then, we have
  $$
(t_i - t_{i-1}) = (s_{k(i-1)} - t_{i-1}) + \left( \sum_{k(i-1) < j < k(i)} (s_j - s_{j-1}) \right) + (t_i - s_{k(i)-1}).
$$
  Then,
  $$
\begin{align*}
(t_i - t_{i-1})\tilde{m}_i &= (s_{k(i-1)} - t_{i-1})\tilde{m}_i + \left( \sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})\tilde{m}_i \right) + (t_i - s_{k(i)-1})\tilde{m}_i\\
&\leq (s_{k(i-1)} - s_{k(i-1)-1})\tilde{m}_i + \left( \sum_{k(i-1) < j < k(i)} (s_j - s_{j-1})m_j \right) + (s_{k(i)} - s_{k(i)-1})\tilde{m}_i\\
\end{align*}
$$
  Now, in the first terms on the RHS replace $\tilde{m}_i$ with $(\tilde{m}_i - m_{k(i-1)} + m_{k(i-1)})$. Similarly, in the third term on the RHS replace $\tilde{m}_i$ with $(\tilde{m}_i - m_{k(i)} + m_{k(i)})$. Expand the brackets appropriately, collect the terms together appropriately, and use $s_j - s_{j-1} = 2^{-N}$ to get
  $$
(t_i - t_{i-1})\tilde{m}_i \leq \sum_{k(i-1) \leq j \leq k(i)} (s_j - s_{j-1}) m_j
+ \frac{\tilde{m}_i - m_{k(i-1)}}{2^N} + \frac{\tilde{m}_i - m_{k(i)}}{2^N}.
$$
  So, by summing over all $0 \leq i \leq n$, we get
  $$
L(f,P,[a,b]) \leq L(f,P_N,[a,b]) + 2 \sum_{i = 0}^n \frac{|m_{k(i)} - \tilde{m}_i|}{2^N}.
$$
  Since $f$ is bounded, $f(x) \leq B$ for all $x \in [a,b]$ for some $B > 0$. Thus, the error is no greater than $4nB/2^N$, which tends to $0$ as $N$ tends to infinity.

A: @Brahadeesh:   The estimate $``\leq (s_{k(i-1)}-s_{k(i-1)-1})\tilde{m}_i+\cdots+(s_{k(i)}-s_{k(i)-1})\tilde{m}_i"$ in Line $-9$ of your foregoing post is not true. Although $t_{i-1}>s_{k(i-1)-1},$ which indeed implies that $s_{k(i-1)}-s_{k(i-1)-1}>s_{k(i-1)}-t_{i-1},$ we can not conclude that
$(s_{k(i-1)}-t_{i-1})\tilde{m}_i\leq (s_{k(i-1)}-s_{k(i-1)-1})\tilde{m}_i,$ because $\tilde{m}_i$ may be negative. Similarly, the final estimate
$``(t_{i}-s_{k(i)-1})\tilde{m}_i\leq (s_{k(i)}-s_{k(i)-1})\tilde{m}_i"$ is not true yet.
