If $f$ is of bounded variation is $f$ Riemann integrable?

I want to know if $f$ is of bounded variation on $[a,b]$ does it follow that $f$ is Riemann integrable on $[a,b]$?

Solution 1:

Here are some useful facts to consider:

• If $f$ is of bounded variation, $f = f_1 - f_2$ with each $f_i$ a monotonically increasing function.

• If $f$ is a monotone function, it can only have jump discontinuities.

• If $f$ has only jump discontinuities, then each discontinuity can be corresponded with a rational number.

Solution 2:

Based on the fact that a function of bounded variation is a difference of two increasing functions, the integrability of functions of bounded variation is reduced to checking that of an increasing function.

Without going into the deep analysis of discontinuities of a monotone function (as suggested in the answer from T. Bongers) we may prove the integrability of a monotone function directly using the upper and lower sums. Suppose $f$ is increasing on $[a, b]$ and $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ is a partition of $[a, b]$ with $x_{k} = a + (k(b - a)/n)$ so that $x_{k} - x_{k - 1} = (b - a)/n$. Then we clearly have $$M_{k} = \sup\,\{f(x)\mid x \in [x_{k - 1}, x_{k}]\} = f(x_{k}),\,\, m_{k} = \inf\,\{f(x)\mid x \in [x_{k - 1}, x_{k}]\} = f(x_{k - 1})$$ and then we get $$U(P, f) - L(P, f) = \sum_{k = 1}^{n}(M_{k} - m_{k})(x_{k} - x_{k - 1}) = \frac{b - a}{n}\sum_{k = 1}^{n} \{f(x_{k}) - f(x_{k - 1})\} = \frac{(b - a)(f(b) - f(a))}{n}$$ and this can clearly be made less than any pre-assigned $\epsilon > 0$ by choosing a suitably large value of $n$. Thus $f$ is integrable on $[a, b]$.