Let $f$ be a continuous, increasing function on $[a,b]$. I know that because $f$ is continuous, I can use epsilon-delta to prove that the lower sum $L(f,P)$ and upper sum $U(f,P)$ both converge given a partition of $[a,b]$: I can divide it equally into $n$ segments, such that the function differs by at most $\frac{\epsilon}{b-a}$ where the specific length of each section is determined by $\delta$ from the epsilon-delta definition. (I think this explanation is correct; if not, please let me know).

However, my main question is does this result hold for a non-continuous strictly increasing function on $[a,b]$? If so, how would we prove that?

Additionally, is there an explicit way to construct a partition (based on $\epsilon, b, a$) such that it is always the case for any strictly increasing function that the lower and upper sum get within epsilon of each other?


An increasing function has at worst a countable number of discontinuities where right- and left-hand limits exist.

Take a uniform partition $P_n = (x_0,x_1, \ldots, x_n) $ with $x_k = a + (b-a)k/n$ and let $f_-(x)$ and $f_+(x)$ denote the left- and right-hand limits. Of course if $f$ is continuous at $x$, then $f_-(x) = f_+(x)$.

Allowing for discontinuities at partition points we have $f_-(x_k) \leqslant f(x_k) \leqslant f_+(x_k)$ where the value $f(x_k)$ could be taken to be anywhere in $[f_-(x_k), f_+(x_k)]$ without affecting the integral.

It is always true that $\sup_{x \in [x_{k-1},x_k]}f(x) = f(x_k)$ and $\inf_{x \in [x_{k-1},x_k]}f(x) = f(x_{k-1})$ regardless of how values are defined at discontinuity points.


$$U(P_n,f) - L(P_n,f) = \frac{b-a}{n}\sum_{k=1}^nf\left(x_k \right) - \frac{b-a}{n}\sum_{k=1}^nf\left(x_{k-1} \right) \\ = \frac{b-a}{n} \left(\sum_{k=1}^n [f(x_k) - f(x_{k-1})] \right) \\ = \frac{b-a}{n} \left(f(b) - f(a) \right), $$

where the last equality follows because the sum is telescoping.

Now observe that $U(P_n,f) - L(P_n,f) \to 0$ as $n \to \infty$ and for all $n > \frac{(b-a)(f(b)-f(a))}{\epsilon}$ we have

$$U(P_n,f) - L(P_n,f) < \epsilon$$


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