I know that the Lebesgue differentiation theorem states that, for every function $f\in \mathcal{L}^1(\mathbb{R}^n)$, the set of its Lebesgue points is almost all $\mathbb{R}^n$ which means that $f$ is approximately continuous for almost every $x\in\mathbb{R}^n$. The thing is: is there a way of proving the particular case
Prop: If $f\in {\mathcal{L}}^1(\mathbb{R})$, then $\exists x_0\in\mathbb{R}:\lim\limits_{x\to x_0}\frac{1}{x-x_0}\int_{x_0}^xf(t)d\lambda(t)=f(x_0)$
without explicitly making use of the LDT? Clearly, if $f$ were locally continuous at some $x_0\in\mathbb{R}$, then $x_0$ would satisfy the desired condition. However, it could so happen that $f$ is nowhere continuous while being Lebesgue integrable, e.g. the Dirichlet function $f=\chi_\mathbb{Q}$. So how could we prove it in general without the said theorem? Thanks :)