Bochner Integral: Lebesgue Point

Question

This thread is just a note.

Given an euclidean space and a Banach space.

Consider Bochner integrable functions: $$F\in\mathcal{B}(\mathbb{R}^d,E):\quad\int\|F\|\mathrm{d}\lambda<\infty$$

Then almost every point is a Lebesgue point: $$D_r(F;z):=\frac{1}{\lambda(B_r(z))}\int_{B_r(z)}F\mathrm{d}\lambda=F(z)$$ How to obtain this from Lebesgue's version?

Answer 1

This is taken from Bochner's original paper.

By strong measurability there are simple functions: $$S_n\in\mathcal{S}(\mathbb{R}^d,E):\quad S_n\to F$$

So one can decompose the above into: $$\|D_r(F;z)-F(z)\|\leq D_r(\|F-S_N\|;z)+\|D_r(S_n;z)-S_N(z)\|+\|F(z)-S_N(z)\|$$

By linearity the second term reduces to: $$D(a\chi_A+b\chi_B;z)=aD(\chi_A;z)+bD(\chi_B;z)$$

So the first and second term can be handled by Lebesgue's version.