If a derivative $f^\prime$ is of bounded variation then prove that $f^\prime$ is continuous. I am having trouble with the following problem:


Let $f:[a,b]\to \mathbb R$ be  differentiable on $[a,b]$ and $f^\prime$ is of bounded variation on $[a,b]$. Prove that $f^\prime$ is continuous on $[a,b]$.


My guess is affirmative. I know that $f^\prime$ can be expressed as difference of two monotonic functions $f^\prime=g-h$ on $[a,b]$. Since $f^\prime$  satisfies intermediate value property (i.e. for any $\lambda$ such that $f^\prime(x)<\lambda<f^\prime(y)$ then there exists a $t$ in $[a,b]$ between $x$ and $y$ such that $f^\prime(t)=\lambda$). From this I am attempting to show that since $f^\prime=g-h$ and both of $g$ and $h$ are monotone, then $f^\prime$ will be continuous. 
But I do not make it. I have seen a nearly similar type of questions in the MSE but I find hard to address it. Let me know whether my guess is right and if so how to proceed further. Thanks for your attention.
 A: For a monotonic function, the left-hand limit and the right-hand limit both exist at each point [except at the endpoints of the interval, where of course only one of the one-sided limits exists]. Hence the one-sided limits also exist everywhere for a difference of monotonic functions, that is, for functions of bounded variation.
Thus a function of bounded variation can have only jump discontinuities, but a function with jump discontinuities doesn't have the intermediate value property, so a derivative of [locally] bounded variation is continuous.

Original overcomplicated answer:
Since $g$ and $h$ are monotonic (let's say monotonically increasing), they both can have at worst jump discontinuities, and at most countably many.
At the points where $g$ and $h$ are both continuous, the difference $g-h$ is also continuous.
So let's look at a point $x\in [a,b]$ where $g$ has a discontinuity. Let
$$\rho_g = \inf \{ g(y) : y > x\} - g(x)\quad \text{and}\quad \lambda_g = g(x) - \sup \{g(y) : y < x\},$$
and define $\rho_h,\lambda_h$ analogously (if $x = a$ set $\lambda_g = \lambda_h = 0$, and for $x = b$, set $\rho_g = \rho_h = 0$).
If $\rho_g \neq \rho_h$, then
$$g(x) - h(x) \neq \lim_{\delta \searrow 0} \bigl(g(x+\delta) - h(x+\delta)\bigr)$$
and $(g - h)\lvert_{[x,b]}$ has a jump discontinuity at $x$, which means that $g-h$ doesn't have the intermediate value property and cannot be a derivative. The same argument shows that $\lambda_g = \lambda_h$ if $g-h = f'$. Now let
$$m(y) = \begin{cases}\quad 0 &, y < x \\ \quad\lambda_g &, y = x\\ \lambda_g + \rho_g &, y > x\end{cases}$$
and $\tilde{g} = g - m,\; \tilde{h} = h - m$. Then $\tilde{g}$ and $\tilde{h}$ are continuous at $x$ and $f' = \tilde{g} - \tilde{h}$, which shows that $f'$ is continuous at $x$.
