I'm not sure if the following statement is true or not, but I intuitively I think it is, I was wondering if I could get some pointers on how to prove it if it is in fact true:
Suppose that $f$ is integrable on $[a,b]$. Then $$ \inf\{U(f,P)\} = \inf\{U(f,Q)\}, $$ where $P$ is an arbitrary partition of $[a,b]$, $Q = P\cup\{c\}$, and $c$ is an arbitrary point between $a$ and $b$.
Yes, equality holds.
Let $S_1:=\text{All partitions on}\ [a,b]$ and $S_2:= \text{All partitions on}\ [a,b] \ \text{with atleast 3 elements in it.}$ Then $S_1=\{(a,b)\} \cup S_2\ $ i.e $S_2 \subset S_1$ $$\implies \operatorname{inf}\{U(f,P):P \in S_1\} \leq \operatorname{inf}\{U(f,P):P \in S_2\}$$
If $P \in S_1$ then $P=(a,b) \ \text{or} \ P \in S_2$. Clearly $\{U(f,P):P \in S_1\}$ contains elements less than $U(f,(a,b))$. So definitely inf of this set is lower than $U(f,(a,b))$. Using $S_1=\{(a,b)\} \cup S_2\ $, every element of $\{U(f,P):P \in S_1\}$ can be bounded from below by an element of $\{U(f,P):P \in S_2\}$ which in turn is bounded by $\operatorname{inf}\{U(f,P):P \in S_2\}$. So we obtain $$\implies \operatorname{inf}\{U(f,P):P \in S_1\} \geq \operatorname{inf}\{U(f,P):P \in S_2\}$$ Hence equality.
Also in general $\operatorname{Sup}\{U(f,P):P \in S_1\} \neq \operatorname{Sup}\{U(f,Q):Q \in S_2\}$.
Consider the function $f:[0, 2] \rightarrow \mathbb{R}$ defined as $$f(x) := \begin{cases} x & 0 \leq x \leq 1 \\ 0 & 1<x \leq 2 \\ \end{cases}$$