Can we conclude the equality of average from Lebesgue point?

Question

Assume $x$ is a Lebesgue point of function $f$. Obviously, we have the following, $$\lim_{r\to 0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|dy=0.$$ Based on this, can we conclude the following equality? $$|f(x)| = \lim_{r\to 0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)|dy.$$ I guess we may need the triangle inequality to prove this.

Answer 1

Note that $|f(y) - f(x)| \geq \big||f(y)| - |f(x)|\big|$. So we have by linearity, monotonicity and absolute value relations: $$ \left[\frac 1{m(B(x,r))} \int_{B(x,r)} |f(y) - f(x)|dy\right] \\\geq \left[\frac 1{m(B(x,r))} \int_{B(x,r)} \big||f(y) - f(x)|\big|dy\right] \\\geq \left[\frac 1{m(B(x,r))} \left|\int_{B(x,r)} |f(y)| - |f(x)|dy\right|\right] \\\geq \left|\frac 1{m(B(x,r))} \int_{B(x,r)} |f(y)| dy - \frac 1{m(B(x,r))} \int_{B(x,r)} |f(x)| dy \right| \\\geq \left|\frac 1{m(B(x,r))} \int_{B(x,r)} |f(y)| dy - |f(x)|\right| $$

First expression goes to zero, then last expression goes to zero. Note that the converse is not true.

Answer 2

Yes. You want to show that $\frac 1 {m(B(x,r)}\int_{B(x,r)} [|f(y)|-|f(x)|]dy \to 0$ and this follows from theinequality $||f(y)|-|f(x)|| \leq |f(y)-f(x)|$.

[I have used the fact that $\frac 1 {m(B(x,r)}\int_{B(x,r)} |f(x)|dy =|f(x)|$].