User:Tkojar/Sandbox/Lebesgue Differentiation Theorem

Statement
For a Lebesgue integrable real or complex-valued function f on $\mathbb{R}^{n}$, the indefinite integral is a set function which maps a measurable set $A$ to the Lebesgue integral of $f \cdot \mathbf{1}_A$, where $\mathbf{1}_{A}$ denotes the characteristic function of the set A.

It is usually written


 * $\displaystyle A \mapsto \int_{A}f\ \mathrm{d}\lambda,$

with $\lambda$ the n-dimensional Lebesgue measure. The derivative of this integral at x is defined to be


 * $\displaystyle \lim_{B \rightarrow x} \frac{1}{|B|} \int_{B}f \, \mathrm{d}\lambda,$

where $|B|$ denotes the volume (i.e., the Lebesgue measure) of a ball B centered at x, and $B\to x$ means that the diameter of $B$ tends to 0.

The Lebesgue differentiation theorem states that this derivative exists and is equal to f\paren{x} at almost every point $x\in \mathbb{R}$.

In fact a slightly stronger statement is true. Note that:


 * $\displaystyle \left|\frac{1}{|B|} \int_{B}f(y) \, \mathrm{d}\lambda(y) - f(x)\right| = \left|\frac{1}{|B|} \int_{B}(f(y) - f(x))\, \mathrm{d}\lambda(y)\right| \le \frac{1}{|B|} \int_{B}|f(y) -f(x)|\, \mathrm{d}\lambda(y).$

The stronger assertion is that the right hand side tends to zero for almost every point x. The points x for which this is true are called the Lebesgue points of $f$.

Proof
Since the statement is local in character, $f$ can be assumed to be zero outside some ball of finite radius and hence integrable.

It is then sufficient to prove that the set


 * $ \displaystyle E_\alpha = \Bigl\{ x \in \mathbf{R}^n :\limsup_{|B|\rightarrow 0, \, x \in B} \frac{1}{|B|} \bigg|\int_B f(y) -f(x)\, \mathrm{d}y\bigg|  > 2\alpha \Bigr\}$

has measure 0 for all $\alpha>0$.

Let $\epsilon>0$ be given. Using the density of continuous functions of compact support in $L^{1}(\mathbb{R})$, one can find such a function $g$ satisfying


 * $\displaystyle\|f - g\|_{L^1} = \int_{\mathbf{R}^n} |f(x) - g(x)| \, \mathrm{d}x < \varepsilon.$

It is then helpful to rewrite the main difference as


 * $\displaystyle \frac{1}{|B|} \int_B f(y) \, \mathrm{d}y - f(x) = \paren{\frac{1}{|B|} \int_B \paren{f(y) - g(y)} \, \mathrm{d}y } + \paren{\frac{1}{|B|}\int_B g(y) \, \mathrm{d}y - g(x) }+ \paren{g(x) - f(x)}.$

The first term can be bounded by the value at x of the maximal function for $f-g$, denoted here by $(f-g)^*(x)$:


 * $\displaystyle \frac{1}{|B|} \int_B |f(y) - g(y)| \, \mathrm{d}y \leq \sup_{r>0} \frac{1}{|B_r(x)|}\int_{B_r(x)} |f(y)-g(y)| \, \mathrm{d}y = (f-g)^*(x),$

which is also called the Hardy–Littlewood maximal function.

The second term disappears in the limit since g is a continuous function, and the third term is bounded by $|f(x)- g(x)|$.

For the absolute value of the original difference to be greater than $2\alpha$ in the limit, at least one of the first or third terms must be greater than $\alpha$ in absolute value.

However, the estimate on the Hardy–Littlewood function (see Hardy–Littlewood maximal inequality) says that


 * $\displaystyle \Bigl| \left \{ x : (f-g)^*(x) > \alpha \right \} \Bigr| \leq \frac{A_n}{\alpha} \, \|f - g\|_{L^1} < \frac{A_n}{\alpha} \, \varepsilon,$

for some constant $A_{n}$ depending only upon the dimension n. The Markov's Inequality (also called Tchebyshev's inequality) says that


 * $ \Bigl|\left\{ x : |f(x) - g(x)| > \alpha \right \}\Bigr| \leq \frac{1}{\alpha} \, \|f - g\|_{L^1} < \frac{1}{\alpha} \, \varepsilon$

whence


 * $\displaystyle |E_\alpha| \leq \frac{A_n+1}{\alpha} \, \varepsilon.$

Since $\epsilon$ was arbitrary, it can be taken to be arbitrarily small, and the theorem follows.