Lebesgue points are defined by: $$\lim_{r \to 0}\frac{1}{Leb(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|dy=0 \text{ for a.e. } x \in \mathbb{R} \text{ w.r.t. } Leb$$
where $f\in L^1(\mathbb{R}), B(x,r)$ is the ball with center $x$ and radius $r$.
My question is: if we replace $Leb$ by general measure $\mu$? Do we still have the same result?
There are results of the type you mention for Radon measures on $\mathbb{R}^d$. There are analytic methods (Besicovitch's coverings, Hardy-Littlewood maximal inequalities) to approach them as well as probabilistic methods (Martingales). Here is one such result:
Theorem:
Suppose $\nu$ and $\mu$ are Radon measure on $\mathbb{R}^d$, $\mu\geq0$. Let \begin{aligned} \lim_{r\rightarrow0}\frac{\nu(\overline{B(x;r)})}{\mu(\overline{B(x;r)})}=D_\mu\nu(x) \end{aligned}
Suppose the Radon--Nikodym decomposition of $\nu$ with respect to $\mu$ is \begin{aligned} \nu=\nu_a+\nu_\perp =\frac{d\nu_a}{d\mu}\cdot\mu + \nu_\perp \end{aligned} be the Radon--Nikodym decomposition of $\nu$ with respect to $\mu$. Then
(i). $D_\mu\nu_a$ exists $\mu$--a.s. and
\begin{aligned} f= \frac{d\nu_a}{d\mu}= D_\mu\nu = D_\mu\nu_a \qquad \text{$\mu$--a.s.}, \end{aligned} and for $\mu$--a.a. $x\in \mathbb{R}^d$, \begin{aligned} \lim_{r\rightarrow0}\frac{1}{\mu\big(\overline{B(x;r)}\big)}\int_{\overline{B(x;r)}}|f(y)-f(x)|\,\mu(dy). \end{aligned} (ii). $D_\mu\nu_\perp$ exists $\mu$--a.s. and \begin{align} D_\mu\nu_\perp=0\qquad\text{$\mu$--a.s.} \end{align}