I have the following integral: $$ \int_0^t f(t)dv(t), $$ where $f(\cdot)$ is continuous and $v(\cdot)$ is piecewise continuously differentiable - so it may have jump discontinuities. Assume for the sake of simplicity that $v(t)$ is right-continuous.
Suppose that $v(\cdot)$ contains a jump at $\tau\in (0,T)$. So it may be decomposed via Lebesgue decomposition into: $$ v(t) = \tilde v(t)+H(t-\tau)(v(\tau)-v(\tau^-)), $$ where $\tilde v(\cdot)$ is continuously differentiable and $H$ is the Heaviside step function. I would like to know why is the above integral then equal to: $$ f(\tau)(v(\tau)-v(\tau^-))+\int_0^t f(t)d\tilde v(t)? $$
My main concern is with the first "jump" term that appears. Intuitively I get it: it's like integrating the constant value $f(\tau)$ over the length of the "gap" produced by the jump. Can someone provide a more rigorous explanation of this? Note that the more general context of this is when $v(\cdot)$ may be just bounded variation - so may contain also "removable discontinuities".