Exercise 1.6.16 (Lebesgue differentiation theorem in general dimension for continous functions)

Show that Theorem 1.6.19 holds for continuous functions.

Question

Show that Theorem 1.6.19 holds for continuous functions.

Elu Thingol · Accepted Answer

By definition,
    $$\forall \epsilon > 0 \quad \exists \delta_\epsilon >0 \quad \forall y \in X: \quad \|y-x\| \leq \delta \implies |f(y) - f(x)| \leq \epsilon $$
    Thus, choosing an appropriate $\delta>0$ we control the size of the integral:
    $$\frac{1}{m(B(x,\delta))} \int_{B(x,\delta)} |f-f(x)| \,\mathrm{d}m \leq \frac{1}{m(B(x,\delta))} \int_{B(x,\delta)} \epsilon \,\mathrm{d}m \leq \frac{1}{m(B(x,\delta))} \cdot {m(B(x,\delta))} \cdot \epsilon = \epsilon  $$

Exercise 1.6.16 (Lebesgue differentiation theorem in general dimension for continous functions)

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Exercise 1.6.16 (Lebesgue differentiation theorem in general dimension for continous functions)

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