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?
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.