Lebesgue Differentiation Theorem

• Jul 30th 2013, 05:35 AM
Kramer
Lebesgue Differentiation Theorem
The Lebesgue Differentiation Theorem states:

If $\displaystyle f: \mathbf{R}^{n} \rightarrow C$ is a summable function then for almost every $\displaystyle x \in \mathbf{R}^{n}$ the following limit converges to $\displaystyle 0$:

$\displaystyle lim_{r \rightarrow 0}\text{ } (\frac{1}{m(B(x,r)})\int_{B(x,r)}|f(y) - f(x)|dy = 0$

Where $\displaystyle m$ is the lebesgue measure and $\displaystyle B(x,r)$ is an open ball of radius $\displaystyle r$.

How does it follow that this Theorem is satisfied if $\displaystyle f$ is only assumed to be continuous?

Let me know if anything is unclear. Thanks. (Headbang)
• Jul 30th 2013, 10:05 AM
Phantasma
Re: Lebesgue Differentiation Theorem
I believe it is required that $\displaystyle f$ be Lebesgue integrable.

See the proof here.
• Aug 8th 2013, 12:09 PM
Rebesques
Re: Lebesgue Differentiation Theorem
Quote:

Originally Posted by Kramer
The Lebesgue Differentiation Theorem states:

If $\displaystyle f: \mathbf{R}^{n} \rightarrow C$ is a summable function then for almost every $\displaystyle x \in \mathbf{R}^{n}$ the following limit converges to $\displaystyle 0$:

$\displaystyle lim_{r \rightarrow 0}\text{ } (\frac{1}{m(B(x,r)})\int_{B(x,r)}|f(y) - f(x)|dy = 0$

Where $\displaystyle m$ is the lebesgue measure and $\displaystyle B(x,r)$ is an open ball of radius $\displaystyle r$.

How does it follow that this Theorem is satisfied if $\displaystyle f$ is only assumed to be continuous?

When f is (additionaly) continuous, then it is uniformly continuous over the closed balls $\displaystyle \overline{B(x,r)}, r>0$. Choose such a ball of radius $\displaystyle r_0>0$. Then, due to uniform continuity, for every $\displaystyle \epsilon>0$ there is $\displaystyle 0<r_1<r_0$ such that $\displaystyle |f(x)-f(y)|<\epsilon , \forall y\in B(x,r_1).$
This implies

$\displaystyle \frac{1}{m(B(x,r_1))}\int_{B(x,r_1)}|f(y) - f(x)|dy \leq \epsilon$

from where the result follows.