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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?

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    $\begingroup$ Maybe define $\nu(A): = \int_A f d\mu$ as a new measure and apply the density theorem to that?! $\endgroup$ Mar 6, 2020 at 1:03
  • $\begingroup$ How do you know that density theorem holds for that measure? $\endgroup$
    – Bob
    Mar 6, 2020 at 3:12
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    $\begingroup$ We are assuming that density theorem holds for all probability measures, right? $\endgroup$ Mar 6, 2020 at 3:20
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    $\begingroup$ No, just for $\mu$. However your comment solved another problem of mine, so thx anyway $\endgroup$
    – Bob
    Mar 6, 2020 at 3:32
  • $\begingroup$ It feels like under some constraints on $(X,d)$, this may be the consequence of en.wikipedia.org/wiki/Lusin%27s_theorem, but not in general. $\endgroup$
    – P. Quinton
    Jul 28, 2020 at 14:52

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