Lebesgue points

Real and Complex Analysis, Page 138

$\textbf{7.6 Lebesgue points}$
If $f\in L^1(R^k)$, any $x\in R^k$ for which it is true that

$\lim_{r\to 0}\frac{1}{m(B_r)}\int_{B(x,r)}|f(y)-f(x)|\,dm(y)=0$

is called a Lebesgue point of $f$.

$|\int_{B(x,r)}(f(y)-f(x))\,dm|\le\int_{B(x,r)}|f(y)-f(x)|\,dm$
$\frac{1}{m(B_r)}|\int_{B(x,r)}(f(y)-f(x))\,dm|\le\frac{1}{m(B_r)}\int_{B(x,r)}|f(y)-f(x)|\,dm$

$\lim_{r\to 0}\frac{1}{m(B_r)}|\int_{B(x,r)}(f(y)-f(x))\,dm|\le\lim_{r\to 0}\frac{1}{m(B_r)}\int_{B(x,r)}|f(y)-f(x)|\,dm=0$

$\lim_{r\to 0}|\frac{1}{m(B_r)}\int_{B(x,r)}f(y)\,dm-f(x)|=0$

(1)

$\lim_{r\to 0}(\frac{1}{m(B_r)}\int_{B(x,r)}f(y)\,dm-f(x))=0$
$\because f(x)是常数，\therefore f(x)=\lim_{r\to 0}\frac{1}{m(B_r)}\int_{B(x,r)}f(y)\,dm$

(2)
$\forall \epsilon>0,\exists \delta>0$使得$\forall r\in (0,\delta)$都有

$|\frac{1}{m(B_r)}\int_{B(x,r)}f(y)\,dm-f(x)|<\epsilon$