# Minumum value of integral and of partition are eventually the same?

I am trying to compare the minimum value of the integral of some function $f$ over the interval $[a,b]$ to the minimum value of the sum of the rectangles up to a certain point in a partition of $[a,b].$ I'm not assuming $f$ continuous here.

Let $P = \{ t_{0},\ldots,t_{n}\}$ be a general partition of $[a,b]$ where we have $t_{0} = a,t_{n} = b.$ Then we can consider the number $$\min\limits_{k=1,\ldots,n} \sum_{i=1}^{k} f(t_{i-1})(t_{i}-t_{i-1})$$ It seems pretty clear that this should approach the value of $$\min\limits_{x \in [a,b]} \int_{a}^{x} f(t) \, dt$$ as the partition gets finer. How do I prove it?

My work

Well the minimum of the integral is achieved at some $x_{0} \in [a,b]$ and we can always choose to have $x_{0} = t_{\mu} \in P.$ Then I have tried doing things like \begin{align*} & \left| \int_{a}^{x_{0}} f(t)\,dt - \min\limits_{k=1,\ldots,n} \sum_{i=1}^{k} f(t_{i-1})(t_{i}-t_{i-1}) \right| \\ \leq {} & \left| \int_{a}^{x_{0}} f(t)\,dt - \sum_{i=1}^{\mu} f(t_{i-1})(t_{i}-t_{i-1}) \right| + \left| \sum_{i=1}^{\mu} f(t_{i-1})(t_{i}-t_{i-1}) - \min\limits_{k=1,\ldots,n} \sum_{i=1}^{k} f(t_{i-1})(t_{i}-t_{i-1}) \right| \\ < {} & \epsilon + \left| \sum_{i=1}^{\mu} f(t_{i-1})(t_{i}-t_{i-1}) - \min\limits_{k=1,\ldots,n} \sum_{i=1}^{k} f(t_{i-1})(t_{i}-t_{i-1}) \right| \end{align*} for an appropriately fine partition $P.$ But now this difference on the right is where I get stuck. It seems so obvious that it should go to zero but I really can't figure out how to prove it. I feel like I'm missing something obvious.

• Are you familiar with the Euler-Maclaurin summation formula? en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula Commented Apr 26, 2014 at 19:22
• Not yet! This hasn't been covered in my text so far and so I think that it ought not to be necessary (provided the author has been careful, but it's Spivak so I think it's a safe bet). I will definitely read the article and try and use it, but I do wonder if there is another way...
– Shai
Commented Apr 26, 2014 at 19:28

For every partition $P,$ assume that the element $x_{o} = t_{\mu} \in P.$ I contend that $$\exists \delta>0 : t_{i} - t_{i-1} < \delta \Rightarrow \sum_{i=1}^{\mu}f(t_{i-1})(t_{i}-t_{i-1}) \leq \min\limits_{k=1,\ldots,n} \sum_{i=1}^{k} f(t_{i-1})(t_{i}-t_{i-1}) + \epsilon$$ for arbitrary $\epsilon>0.$
Suppose, for contradiction, that this is not true. Then there is some $d>0$ such that, for any $\delta>0,$ we can always find a partition $P$ with $t_{i}-t_{i-1}<\delta$ containing some element $t_{m} \neq t_{\mu}$ satisfying $$\sum_{i=1}^{\mu}f(t_{i-1})(t_{i}-t_{i-1}) > \sum_{i=1}^{m} f(t_{i-1})(t_{i}-t_{i-1}) + d.$$ Well, then let $\delta>0$ be so small that for any partition $P$ with $t_{i}-t_{i-1}<\delta$ we have $$\left| \int_{a}^{b} f - \sum_{i=1}^{n} f(t_{i-1})(t_{i}-t_{i-1}) \right| < \frac{d}{4}$$ It follows (from a theorem contained in the book) that for any such partition, we also have $$\left| \int_{a}^{x_{0}} f - \sum_{i=1}^{\mu} f(t_{i-1})(t_{i}-t_{i-1}) \right|, \left| \int_{a}^{t_{m}} f - \sum_{i=1}^{m} f(t_{i-1})(t_{i}-t_{i-1}) \right| < \frac{d}{2}$$ Then we are forced to conclude that \begin{align*} \int_{a}^{x_{0}} f - \int_{a}^{t_{m}} f & > \left( \sum_{i=1}^{\mu} f(t_{i-1})(t_{i}-t_{i-1}) - \frac{d}{2} \right) - \left( \sum_{i=1}^{m} f(t_{i-1})(t_{i}-t_{i-1}) + \frac{d}{2}\right) \\ & > d-d \\ & = 0, \end{align*} which contradicts the minimality of $\int_{a}^{x_{0}} f.$