Concerning the integrability of a bounded, monotonic function.

Question

Here is a question that's been bothering me:

Suppose $f$ is a monotonic and bounded (say, real-valued) function on $[a, b) \subset \mathbb{R}$, and that $f(b) = f(a)$. (Let's also assume that $f$ is not constant.)

Is $f$ Riemann-integrable on $[a, b]$?

My thought is that, since $f$ is bounded on $[a,b]$, it should be integrable by Lebesgue's Theorem—that is, if its set of discontinuities in that interval is of measure zero (which seems plausible, but I am having trouble proving this).

Answer 1

Well, $f$ is bounded and monotonic on $[a, b) $ and hence $L=\lim_{x\to b^{-}} f(x) $ exists. Define $f(b) =L$ and then $f$ is monotonic on $[a, b] $ and hence Riemann integrable on $[a, b]$.

The function $f$ in your question differs from the one mentioned in last paragraph only due the value taken at $b$ (you have $f(b) =f(a) $ which may or may not be equal to the number $L$ used in last paragraph). Now remember that changing the values of a function at a finite number of points does not change its Riemann integrability or the value of its integral (provided the integral exists) and hence your function is also Riemann integrable on $[a, b] $.

There is no need to invoke the Lebesgue's criterion of Riemann integrability and a simpler analysis as presented above suffices for most problems dealing with Riemann integration.

Answer 2

I think the question that you are concerning is the following:

If $f$ is an increasing/nondecreasing function on $[a,b]$, then $f$ is Riemann integrable on $[a,b]$.

The boundedness is no needed to be emphasized, because that $f$ being increasing entails that $f(a)\leq f(x)\leq f(b)$ for all $x\in[a,b]$.

Now we prove the Riemann integrability. Given $\epsilon>0$, take a partition $P=\{a=x_{0},...,x_{n}=b\}$ with $\max_{1\leq i\leq n}(x_{i}-x_{i-1})<\epsilon$, we have \begin{align*} U(f,P)-L(f,P)&=\sum_{i}\left(\sup_{i\in[x_{i-1},x_{i}]}f(x)-\inf_{i\in[x_{i-1},x_{i}]}f(x)\right)\Delta x_{i}\\ &\leq\sum_{i}(f(x_{i})-f(x_{i-1}))\Delta x_{i}\\ &\leq\sum_{i}(f(x_{i})-f(x_{i-1}))\epsilon\\ &=(f(b)-f(a))\epsilon. \end{align*}

Answer 3

Any function that is bounded on $[a,b],$ and is Riemann integrable (RI) on $[a,c]$ for $a<c<b,$ is RI on $[a,b].$ Your function is bounded on $[a,b]$ and is monotonic on $[a,c]$ for $a<c<b.$ Hence it is RI on each such $[a,c].$ Your result follows.