On Wikipedia I find the following version of the integral mean value theorem (MVT) for continuous functions:
If $f\colon[a,b]\to\mathbb{R}$ is continuous, then there exists a $c\in(a,b)$ such that $$\int_{a}^{b}f(t) \ \text{d}t=(b-a)f(c).$$
Now let $f\colon[a,b]\to\mathbb{R}$ be a map with bounded variation. This means that $$\sup_{a=x_{1}<x_{2}<\ldots<x_{n}=b}\sum_{j=1}^{n-1}|f(x_{j})-f(x_{j+1})|<\infty.$$ According to this link (proposition 5) the set of points of discontinuity of $f$ is at most countable. Does the integral MVT still hold?
No: there are even functions with one discontinuity for which this is false. For example, take the function $f$ which is $1$ for positve arguments and $-1$ otherwise. Then $f$ is nondecreasing, so has bounded variation, and never takes the value $0$, but $$ \int_{-1}^{1} f = 0. $$