What is the connection between the Radon-Nikodym derivative and the fundamental theorem of calculus in $mathbb{R}^n$?

Section 3.4 of Folland’s Real Analysis is titled "Differentiation on Euclidean space". In it he presents material that he says "may be regarded as a generalization of the fundamental theorem of calculus". I think I understand the connection in the one-dimensional case, but not in higher dimensions. (The Folland section is written in terms of balls in $mathbb{R}^n$, not intervals in $mathbb{R}$.) Can anyone explain that latter connection and/or point out mistakes/misconceptions in the below?

In one dimension ($mathbb{R}$) under Lebesgue measure $m$, the connection is this:

Suppose we have a measure $nu$ and the Radon-Nikodym derivative w.r.t. $m$ exists. Then for an interval $[x, x + r]$ we have
begin{align*}
nu([x,x + r] = int_{x}^{x+r} f dm.
end{align*}
But this integral can be regarded as the difference between two integrals:
begin{align*}
nu([x, x + r]) = int_0^{x + r} f dm – int_0^{x} f dm,
end{align*}
and thus $f^* = lim_{r to 0}frac{nu([x,x + r])}{r}$ is a derivative of the function $F(x) = int_0^x f dm$.

Folland theorem 3.18 states that $lim_{r to 0} A_r f(x) = f(x)$ for a.e. $x in mathbb{R}^n$, where $A_r f(x)$ is the average value of $f$ on $B(x, r)$.

We also note that $nu([x, x + r])$ is the average value of $f$ on the interval, so then the point of Folland theorem 3.18 is that it tells us that $f^* = f$ a.e. which is an FTC-like statement: in simple terms, the derivative of the "area-so-far​" function $F$ is $f$ itself.

Now, how does this extend to $mathbb{R}^n$ for $n>0$? We have
begin{align*}
f^* = lim_{r to 0} frac{nu(B(x, r))}{m(B(x, r))},
end{align*}
and $nu(B(x, r)) = int_{B(x, r)} f dm$, but I don’t see how to make a statement analogous to
begin{align*}
nu([x, x + r]) = int_0^{x + r} f dm – int_0^{x} f dm,
end{align*}
regarding the ball $B(x, r)$ in $mathbb{R}^n$ for $n > 1$.