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 $[a,b]$? 
 A: 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 $[a, b]$.
A: I don't understand why people always prefer the solution with the representation of a BV function as a difference of two monotone functions. This is of course correct but it needs additional work. A proof with the definition of integrability, in my opinion, is way more convenient and self-contained without using any 'black boxes'. The weird thing is that this proof cannot be found anywhere on the internet as if someone really needs to quote the aforementioned representation...
Let $\varepsilon>0$ and $S(P),$ $s(P)$ be the upper and the lower sum respectively of the your function with respect to the partition $P=\{a=x_0<\ldots<x_n=b\}$ with $x_j-x_{j-1}=\frac{b-a}{n}$ (we will choose this $n$ later). Then, we have that 
$$S(P)-s(P)=\sum_{j=1}^n (M_j-m_j)(x_j-x_{j-1})=\frac{b-a}{n}\cdot\sum_{j=1}^n (M_j-m_j),$$ where $M_j,$ $m_j$ is the sup and the inf of the function at the corresponding subinterval $[x_{j-1},x_j]$ respectively. Pick $c_j, c_j'\in [x_{j-1},x_j]$ with $f(c_j)>M_j-1/n$ and $f(c_j')<m_j+1/n.$ Then, if $K$ is the BV constant, we have that 
$$S(P)-s(P)\leq \frac{b-a}{n}\cdot\sum_{j=1}^n \left( \left|f(c_j)-f(c_j')\right| +2/n\right)<\frac{b-a}{n}\cdot(K+2)<\varepsilon$$ for the appropriate $n.$ We are done by the characterization of the definition of the Riemann integrability. 
A: 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.
A: The answer is yes f is riemann integrable
Since f is of bounded variation on say a, b ],  f can be written as a difference of two monotone increasing functions and since any increasing function on (a, b) is integrable it follows that f is integrable by sum difference formula
