Questions about integrability on $\mathbb{R}$

Question

My textbook states the following:
i) If $ f : [a,b] \rightarrow \mathbb{R} $ is bounded and is continuous at all but finitely many points of $[a,b]$, then it is integrable on $[a,b]$.
ii) Any increasing or decreasing function on $[a,b]$ is integrable on $[a,b]$.

The proof for (i) is clear to me. I followed the entirety of it. My issue is with (ii). Is boundedness and continuity not necessary for (ii), or are they somehow implied by being strictly increasing/decreasing?

Answer 1

I suspect that by increasing and decreasing they mean that it's increasing / decreasing on the entire domain, which is $\mathbb{R}$. So it's defined everywhere on $\mathbb{R}$. Since it's defined everywhere, it's bounded on any compact subset of $\mathbb{R}$ (since any real function attains its supremum on any compact subset).

Continuity is not required for integrability; however, it is the case that continuity needs to be satisfied on all but a finite set of points in an interval. More specifically, the set of discontinuous points in the interval would have to have measure 0. I think that this must be the case for a function that is either decreasing or increasing.

Answer 2

Let $P = \{ a = t_0 < \cdots < t_n = b \}$ be any partition on $[a, b]$. If $f : [a, b] \to \Bbb{R}$ is monotone increasing, then the upper Riemann sum is

$$ U(P, f) = \sum_{i=1}^{n} \sup_{t \in [t_{i-1}, t_i]} f(t) (t_i - t_{i-1}) = \sum_{i=1}^{n} f(t_i) (t_i - t_{i-1}) $$

and likewise the lower Riemann sum is

$$ L(P, f) = \sum_{i=1}^{n} \inf_{t \in [t_{i-1}, t_i]} f(t) (t_i - t_{i-1}) = \sum_{i=1}^{n} f(t_{i-1}) (t_i - t_{i-1}). $$

Let us denote $\| P \| = \max_{1 \leq i \leq n} (t_i - t_{i-1})$ the mesh size of $P$. Taking the difference, we get

\begin{align*} U(P, f) - L(P, f) &= \sum_{i=1}^{n} (f(t_i) - f(t_{i-1})) (t_i - t_{i-1})\\ &\leq \sum_{i=1}^{n} (f(t_i) - f(t_{i-1})) \| P \| \\ &= (f(b) - f(a)) \| P \|. \end{align*}

Since we can make $\| P \|$ arbitrarily small, this proves that $f$ is Riemann integrable on $[a, b]$. The argument is analogous for monotone-decreasing case.

Answer 3

Hint: Use the fact that monotone function can have at most countable number of discontinuity points. So the Lebesgue measure of discontinuity points is zero and thus Riemann integrable.