Does the fundamental theorem of calculus hold for BV functions?

Question

I am a bit confused and I hope you can help me in understanding a bit better these things.

Let us start by considering one dimensional case. Let $f\colon \mathbb (a,b) \to \mathbb R$ be a function. As far as I know, the equality $$\tag{FTC} f(x) = f(a) + \int_a^x f^{\prime}(t)dt $$
holds for every $x \in (a,b)$ if and only if the function $f$ ia $AC(a,b)$. This makes sense, because an absolutely continuous function on the real line is differentiable Lebesgue a.e. (for instance, because it can be written as sum of monotone functions: then a deep theorem by Lebesgue assures a.e. differentiability). Moreover, we know that AC is a (proper) subset of BV; the formula (FTC) does not hold in BV (counterexample: the devil staircase).

Am I right so far? Hope so.

Now let turn to the n-dimensional case, $f \colon \mathbb R^n \to \mathbb R$. We define a BV function as a function whose first (distributional) derivatives are signed measures with finite total variation.

Now the point is: here I read (Theorem 9) that for every BV function there is a signed measure such that $f=\mu((-\infty,x])=\int_{-\infty}^x d\mu$. But this is exactly FTC, up to writing in the RHS $\int_{\infty}^x f^{\prime}$ instead of $\int f^{\prime}dt$!

I mean, the only difference is that $\mu$ need not be absolutely continuous (as a measure) with respect to Lebesgue measure.

So my final conjecture is: FTC holds formulation for every BV function but the RHS must be considered as the distributional derivative. And AC functions are precisely the BV functions whose distributional derivative is absolutely continuous wrt Lebesgue.

Is this correct? Is there anything to add? Thanks.