Let $M$ be a real number,
If $1.\ \forall\epsilon\gt0,$ there exists a partition $P$ of $[a,b]$ such that $M-\epsilon \lt L_f(M)\le M.$
$\ \ \ 2.\ L_f(M)\le M$ for every partition P of $[a,b] $ .
How to prove $M=\underline{I}_a^b(f)$
In my opinion, since we can use contradiction to prove the first condition, $M$ has to be least upper bound for the condition to work, but I don't know how to formally prove it. So if $M$ is certainly a least upper bound, then by the second condition, we confirm that $M$ it the least upper bound for all lower sum and thus is the lower integral. Could someone help with the proof?
First some clarification. For a (nonempty) real set, if there is any upper bound at all, there is then a least upper bound (aka supremum), one and only one.
You're trying to take the least upper bound of the set of $L_f(P)$'s over all partitions $P$. Your first condition is that you can find an $L_f(P)$ which is $> M-\epsilon$ for positive $\epsilon$. That means that, for any positive $\epsilon$, the number $M-\epsilon$ is not an upper bound of the set of all $L_f(P)$'s.
The second condition is that $M$ is an upper bound of the set of all $L_f(P)$'s.
Putting the two things together, $M$ is the least upper bound of that set. Anything less than $M$ and it's not an upper bound.