If $S$ is set of lower sums for partitions having $n$-equal subintervals, does $\sup S$ equal the $\sup$ of the set of lower sums for all partitions. Firstly, here are two relevant definitions:

Let $a \lt b$. A partition of the interval $[a,b]$ is a finite collection of points in $[a,b]$, one of which is $a$ and one of which is $b$.


Suppose $f$ is bounded on $[a,b]$ and $P=\{t_0,t_1, \cdots, t_n\}$ is a partition of $[a,b]$. Let \begin{align} m_i=\inf\{f(x):t_{i-1} \leq x \leq t_i\}. \end{align} The lower sum of $f$ for $P$, denoted by $L(f,P)$ is defined as \begin{align}\displaystyle L(f,P) = \sum_{i=1}^n m_i(t_i-t_{i-1})\end{align}


Consider the set $\mathcal S$ of all partitions of $[a,b]$ that have $n$ equal subintervals. For example, if the interval of interest is $[0,4]$, then the partition $P=\{0,1,2,3,4\}$ is in $\mathcal S$...the partition $P=\{0,0.5,1.0,1.5,\cdots,3.5,4\}$ is in $\mathcal S$...etc. Additionally, let the set $\mathcal T$ denote the set of all possible partitions of $[a,b]$.
It can be shown that there are partitions in $\mathcal T$ that are not in $\mathcal S$: i.e. there are partitions that cannot be represented by any partition that has $n$ equal subintervals. This leads me to wonder the following:

Is it true that $\displaystyle \sup_{P \in \mathcal S} L(f,P)=\sup_{P \in \mathcal T} L(f,P)$ $\quad(*)$?

I've played around with this for a while but am having difficulties finding a scheme that works for any function.
In order to prove $(*)$, I have specifically been trying to prove that:

For any $P \in \mathcal T$, there exists a $P' \in \mathcal S$ such that $L(f,P') \geq L(f,P)$.

, which I think is the correct approach. Any suggestions or insights would be appreciated.

Edit: Adding some additional clarifying details to Nico Tripeny's argument
Relevant Lemma:

Let set $A$ and set $B$ have supremums $\alpha$ and $\beta$, respectively. Suppose \begin{align}\dagger_1 \quad &\forall a \in A: \forall \varepsilon \gt 0: \exists b \in B: a \lt b + \varepsilon \\\dagger_2 \quad &\forall b \in B: \exists a \in A: b \leq a \end{align} then we must have that $\alpha = \beta$.

The fact that $\mathcal S \subset \mathcal T$ immediately gives us $\dagger_2$. Now, we merely need to prove $\dagger_1$, which Nico Tripeny's proof does in the following way (I will generalize to $[a,b]$ rather than $[0,1]$):
Consider a partition $P \in \mathcal T$. In order for this argument to not be trivial, suppose there exists a $p_j \in P$ that cannot be expressed as $a+i\cdot\frac{b-a}{n}$ for any $n \in\mathbb N$ and any $i \in \{0,1,\cdots, n-1,n\}$, which is equivalent to saying that $P \notin \mathcal S$. Because $p_j\neq  a+i\cdot\frac{b-a}{n}$, for a large enough $k \in \mathbb N$, I can guarantee that there is partition $P' \in \mathcal S$ that contains consecutive members $q_i$ and $q_{i+1}$ such that $q_i \lt p_j \lt q_{i+1}$.
To see this, suppose $P$ has $n+1$ members ($n$ intervals). Because this number is finite, one can collect all of the differences between between any two consecutive members and find the smallest such difference $d \gt 0$. By the Archimedean principle, there is a $k \in \mathbb N$ such that $\frac{1}{k}\lt \frac{d}{b-a}$, which means that $\frac{b-a}{k} \lt d$. Importantly, this implies that for every subinterval of $P$, denoted by $[p_j, p_{j+1}]$, there is a $q \in P'$ such that that $q \in (p_j,p_{j+1})$.
The following picture will be useful to understanding the argument:

Recalling our definition of lower sums, it is easy to see that for a given $i^{\text{th}}$ subinterval, we can rewrite the contributing sum from $m_i\cdot(t_i-t_{i-1})$ to $m_i\cdot (z-t_{i-1})+m_i\cdot(t_i-z)$ for some $z \in [t_{i-1},t_i]$.
With this in mind and referencing our above photo, consider the contributions that the following partial intervals make to the lower sums of $L(f,P)$ and $L(f,P')$:
$$(p_j-q_i)\cdot m_{p_j,p_{j-1}}+(q_{i+1}-p_j)\cdot m_{p_{j+1},p_j}$$
and
$$(p_j-q_i)\cdot m_{q_{i+1},q_i}+(q_{i+1}-p_j)\cdot m_{q_{i+1},q_i}$$
It should be straightforward to see that $\displaystyle m_{q_{i+1},q_i} \leq m_{p_j,p_{j-1}}$ and $\displaystyle m_{q_{i+1},q_i} \leq m_{p_{j+1},p_j}$. Moreover, though, because $f$ is bounded on $[a,b]$, we must have that:
$$-M \leq \displaystyle m_{q_{i+1},q_i} \leq m_{p_j,p_{j-1}} \leq M \text{ and } -M \leq \displaystyle m_{q_{i+1},q_i} \leq m_{p_{j+1},p_j} \leq M$$
Now, consider the two differences $\mathcal D$ and $\mathcal C$:
$$\mathcal D= (p_j-q_i)\cdot m_{p_j,p_{j_1}}+(q_{i+1}-p_j)\cdot m_{p_{j+1},p_j} - (p_j-q_i)\cdot m_{q_{i+1},q_i}+(q_{i+1}-p_j)\cdot m_{q_{i+1},q_i}\geq 0$$
$$\mathcal C = (p_j-q_i)\cdot M+(q_{i+1}-p_j)\cdot M - (p_j-q_i)\cdot (-M)+(q_{i+1}-p_j)\cdot (-M)\geq 0$$
We can rewrite $\mathcal C$ as $\displaystyle 2M(q_{i+1}-q_i)$, More importantly, $\mathcal C \geq \mathcal D \geq 0$, so we have:
$$2M(q_{i+1}-q_i) \geq \mathcal D \geq 0$$
Obviously, we repeat this argument for any $p \in P$ that cannot be expressed as $a+i\cdot\frac{b-a}{n}$. For our argument, if the $P$ in question has $n+1$ members, then it has $n$ intervals...but our argument is only needed for the first $n-1$ intervals. So, for a given $P$, the worst case scenario is that we need this argument $n-1$ times.
The essence of Nico Trepeny's argument is that $\mathcal D$ can be made appropriately small such that the difference between a given $L(f,P)$ and an appropriately selected $L(f,P')$ can always be made to be within any $\varepsilon \gt 0$ of each other. You can see that the method for reducing $\mathcal D$ is to shrink the interval size  $q_{i+1}-q_i$, which can be done by selecting the appropriate number $k$ of intervals...which is thus the approach for selecting the appropriate $P'$.
Note that for a $P'$ with $k$-equal intervals, $q_{i+1}-q_{i}$ is simply $\frac{b-a}{k}$. Given that there are potential $n-1$ $\mathcal D$'s that need to be taken into account, we have the list of inequalities:
\begin{align}&2M \cdot \frac{b-a}{k} \geq \mathcal D_1 \geq 0\\&2M \cdot \frac{b-a}{k} \geq\mathcal D_2 \geq 0 \\&\cdots \\& 2M \cdot \frac{b-a}{k} \geq\mathcal D_{n-1}\geq 0 \end{align}
Summing across all of these $\mathcal D$'s, we see that the total difference between $L(f,P)$ and $L(f,P')$ will be $(n-1)\cdot 2M \cdot \frac{b-a}{k}$.
$$L(f,P)-L(f,P')\leq (n-1)\cdot 2M \cdot \frac{b-a}{k}$$
For an arbitrary $\varepsilon$, choose the appropriate $k$, thus dictating which $P' \in \mathcal S$ is needed.
 A: Unfortunately I cannot comment because I don’t have enough rep, but here is an idea of why your approach might not be working. Let $f$ from $[0,1]$ be defined so $f(x)=0$ when $x<\frac{1}{\sqrt{2}}$ and $1$ otherwise. The partition $\{0, \frac{1}{\sqrt{2}},1\}$ will give a value of $1-\frac{1}{\sqrt{2}}$. But for any value of $n$, the partitions you discuss will not include $\frac{1}{\sqrt{2}}$ so will always have a smaller value.
This doesn’t mean the statement is false of course, just that the method you suggest won’t work for the general case.
A: For continuous functions (or more generally Riemann integrable functions) it is true that $\sup_{P \in\mathcal{S}}L(P,f) = \sup_{P\in \mathcal {T}}L(P,f)$, where $\mathcal{S}$ is the set of uniform partitions and $\mathcal{T}$ is the set of all partitions.
This follows because for a Riemann integrable function, the supremum of lower sums over all partitions is the integral, viz.
$$\sup_{P \in\mathcal{T}}L(P,f) = \underline{\int}_a^b f(x) \, dx  =  \int_a^b f(x) \, dx $$
Now for any  $L(P,f)\in \mathcal{S}$, the lower sum is bounded above as $L(P,f) \leqslant \int_a^b f(x) \, dx$.  Since such lower sums over uniform partitions converge to the integral as $\|P \| \to 0$ it is easily shown that
$$\sup_{P\in\mathcal{S}}L(P,f) = \int_a^b f(x) \, dx=\sup_{P\in\mathcal{T}}L(P,f)$$
Addendum
This result is, in fact, true for any bounded function.  This follows because it can be shown that for any $\epsilon > 0$ there exists $\delta > 0$ such that if the norm of the partition satisfies $\|P\| < \delta$, then the lower Darboux sum is within $\epsilon$ of the lower Darboux integral, viz.
$$\tag{*}\underline{\int}_a^b f(x) \, dx - \epsilon< L(P,f) \leqslant\underline{\int}_a^b f(x) \, dx$$
A proof of this fact is indicated here.
Since,
$$\underline{\int}_a^b f(x) \, dx = \sup_{P \in \mathcal{T}}L(P,f),$$
we have, using (*), for any uniform partition $P \in \mathcal{S}$ with $\|P\| < \delta$,
$$\sup_{P \in \mathcal{T}}L(P,f)-\epsilon< L(P,f) \leqslant \sup_{P \in \mathcal{S}}L(P,f) \leqslant \sup_{P \in \mathcal{T}}L(P,f)$$
Since $\epsilon$ can be arbitrarily close to zero it follows that $$\sup_{P \in \mathcal{S}}L(P,f) = \sup_{P \in \mathcal{T}}L(P,f)$$
A: Here is a (sketch of a) solution to the question. Assume $f:[0,1]\to\mathbb{R}$ is bounded so that $|f(x)|<M$ for all $x$ in the domain. It suffices to show that for all $P\in\mathcal{T}$ and all $\varepsilon$, there is a $P'\in \mathcal{S}$ such that $L(f,P)<L(f,P')+\varepsilon$. Let $l$ be the number of elements in $P$, then make $n > \frac{2lM}{\varepsilon}$ where $n$ is the number of elements in $P'$.
We see that most intervals in $P'$ will be completely within those of $P$, except for maybe $l$, as this is the number of points in $P$. If the interval is inside one from $P$, then we can ignore it as the value it adds to the sum will be greater then or equal to the what that section adds in $P$. By boundedness, the intervals in $P'$ that intersect those in $P$ can only have an infimum $2M$ smaller than the corresponding part of $P$. Using this and that there are only $l$ such intervals, we see $L(f,P)<L(f,P') + 2lM\frac{\varepsilon}{2lM}= L(f,P')+\varepsilon$.
The basic idea is once the partitions in $\mathcal{S}$ are fine enough, almost all intervals fall within the intervals in $P$ for a general partition. So the values of the lower sum become very close together.
