The precise representative of a function $f\in L^1_{loc}(\mathbb{R^n})$ (as for example in Evans - Measure Theory and Fine Properties of Functions (2nd Ed)) is defined as
$$ f^*(x) = \lim_{r \rightarrow 0^+} \frac{1}{|B(x,r)|} \int_{B(x,r)} f(y)dy $$
where the limit exists, $0$ elsewhere (but it can be shown that the limit exists almost everywhere!).
It is true that if $f = g$ almost everywhere (i.e., $f$ and $g$ belong to the same class in $L^1_{loc}$), then $f^* = g^*$ (it follows the definition of precise representative in the aforementioned Evan's book).
My question is: is the converse also true? ($f^* = g^* \Rightarrow f=g$ a.e.?). If this is the case, I can conveniently think of the classes in $L^1_{loc}$ as represented by the common precise representative of any element in the class.
I tried to give myself an answer. However, the fastest way to see this is definitley, as it has been suggested in the comments, passing through Lebesgue-Besicovitch differentiation theorem, which assures that $f^*=f$ a.e. I am still posting it; please feel free to comment or correct me.
Our thesis is: $f^* = g^* \Rightarrow f=g$ a.e.
Let's suppose for simplicity that $f\ge0$, and $g=0$ for the moment; the thesis becomes $f^*=0 \Rightarrow f=0$ a.e. Let $x$ be an element of $\mathbb{R}^n$ fixed.
For all $\varepsilon>0$ set $ E_x(r,\varepsilon) := \{ y \in B(x,r) \mid f(y) > \varepsilon \} $. Then we have: $$ \frac{1}{|B(x,r)|} \int_{B(x,r)} fdy \ge \frac{1}{|B(x,r)|} \varepsilon |E_x(r,\varepsilon)| \ge 0 $$ And the left term goes to $0$ when $r\rightarrow0^+$, which implies that $|E_x(r,\varepsilon)| = o(r^n)$.
Now, for simplicity set $n=1$ (but the prove doesn't seem to be much harder in the general case). We take some interval $[a,b]$ and we show that $f=0$ a.e. on $[a,b]$. More precisely, setting $G := \{ y \in [a,b] \mid f(y)>0 \}$, let's show that $|G| = 0$.
For every $k \in 2(\mathbb{N}+1)$, let's split $[a,b]$ as $k$ intervals of length $\frac{2}{k}$: $a=x_0<x_1<\cdots<x_{k/2}=b$ such that $[a,b] = \bigcup_{j=1}^k I_j$ (and $|I_j| = \frac{2}{k}$). Set $ G_j := \{ y \in I_j \mid f(y)>0 \} $. Then;
$$ |G| = \sum_{j=1}^{k/2} |G_j| = \sum_{j=1}^{k/2} o\Big(\frac{1}{k}\Big) \rightarrow_{k\rightarrow+\infty} 0 $$
Now, if $n>1$ we pick rectangles instead of intervals.
Finally, for the general case $f^* = g^*$, we got:
$$ 0 = f^* - g^* = \lim_{r\rightarrow 0^+} \frac{1}{|\{B(x,r)\}|} \int_{B(x,r)}\Big(f-g\Big)dy \Rightarrow (f-g)^*=0 \Rightarrow f-g=0 $$