Let $(X,d)$ be a separable complete metric space and $\mu$ a probability measure on the Borel subsets of $(X,d)$. Suppose that the Lebesgue's density theorem holds, i.e. that for each Borel set $A$ of $(X,d)$ it holds that $$\lim_{r\to0^+}\frac{\mu(A\cap\bar{B}_r(x))}{\mu(\bar{B}_r(x))}=1$$ for $\mu$-almost all $x\in A\cap\operatorname{supp}(\mu)$, where $\operatorname{supp}(\mu)=\{x\in X : \forall r>0, \mu(\bar{B}_r(x))>0\}$.
Question: is it true that Lebesgue's differentiation theorem holds in $(X,d,\mu)$, i.e. that for each $f\in L^1(\mu)$, $$\lim_{r\to0^+}\frac{1}{\mu(\bar{B}_r(x))}\int_{\bar{B}_r(x)}f\operatorname{d}\mu=f(x)$$ for $\mu$-almost all $x\in \operatorname{supp}(\mu)$?
I tried to approximate $f$ in $L^1(\mu)$ by a simple function $g$, since it is easy to show that for $g$ simple the theorem holds, so \begin{align*} &\left|\frac{1}{\mu(\bar{B}_r(x))}\int_{\bar{B}_r(x)}f\operatorname{d}\mu-f(x)\right| \le \frac{1}{\mu(\bar{B}_r(x))}\int_{\bar{B}_r(x)}|f-g|\operatorname{d}\mu \\ & \quad\quad\quad\quad + \left|\frac{1}{\mu(\bar{B}_r(x))}\int_{\bar{B}_r(x)}g\operatorname{d}\mu-g(x)\right|+|g(x)-f(x)|, \end{align*} and then the middle term tends to zero for $\mu$-almost all $x\in\operatorname{supp}(\mu)$. Also Tchebychev allows us to treat the term $|f(x)-g(x)|$. But when trying to estimate the first term it seems that we need a maximal function estimate in terms of $L^1(\mu)$ norm. Now, if we work in $\mathbb{R}^d$ we can rely on Besicovitch covering theorem, but in our context this theorem doesn't hold in general. So I'm wondering if there is any other way to attack this problem without assuming any extra hypothesis... Any ideas?