# Math Help - Lebesgue Differentiation Theorem

1. ## Lebesgue Differentiation Theorem

The Lebesgue Differentiation Theorem states:

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

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

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

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

Let me know if anything is unclear. Thanks.

2. ## Re: Lebesgue Differentiation Theorem

I believe it is required that $f$ be Lebesgue integrable.

See the proof here.

3. ## Re: Lebesgue Differentiation Theorem

Originally Posted by Kramer
The Lebesgue Differentiation Theorem states:

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

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

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

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

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

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

from where the result follows.