Follan - Real Analysis p.243 Theorem 8.15
As far as I know, the term, Lebesgue point $x$, is defined for $L^1$ functions such that $\lim_{r\to 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} |f - f(x)| dm =0$.
Let $f$ be of $L^p$. The author wrote that "If $x$ is a Lebesgue point of $f$, then for any $\delta>0$ there exists $\epsilon>0$ such that $\int_{|y|<r} |f(x-y)-f(x)| dy \leq \delta r^n$ for $r\leq \epsilon$. But what does it mean "a Lebesgue point of $f$" in this case? Why is the above inequality true when $f$ is $L^p$?
Notice that for $x \in \mathbb{R}^n$, $f \in L^p(B_1(x)) \subset L^1(B_1(x))$, so we can talk about Lebesgue points for $f$ regarding $f$ an $L^1_{loc}$ fucntion.
To prove the inequality notice that, if $x$ is a Lebesgue point for $f$, then for every $\epsilon > 0$ there is a radius $R$ such that if $r \le R$ then $$\frac{1}{|B_r(x)|}\int_{B_r(x)}|f(z) - f(x)|\,dz < \epsilon.$$
Now multiply both sides by $|B_r(x)|$, let $\delta$ be such that $\delta r^n = \epsilon|B_r(x)|$ and finally change variables in the integral on the LHS (let $u = z + x$).