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Spring 2009 - Problem 5

Lebesgue differentiation theorem

Let $I = I_{0,0} = \br{0, 1}$ be the unit interval, and for $n = 0, 1, 2, \ldots,$ and $0 \leq j \leq 2^n - 1$ let

I_{n,j} = \br{j2^{-n}, \p{j+1}2^{-n}}.

For $f \in L^1\p{I, \diff{x}}$ define

E_nf\p{x} = \sum_{j=0}^{2^n-1} \p{2^n\int_{I_{n,j}} f \,\diff{t}}\chi_{I_{n,j}}\p{x}.

Prove that if $f \in L^1\p{I, \diff{x}}$ then $\lim_{n\to\infty} E_nf\p{x} = f\p{x}$ almost everywhere on $I$ .

Solution.

Let $x$ be a Lebesgue point of $f$ , i.e., a point where the Lebesgue differentiation theorem applies. Because the $\set{I_{n,j}}_j$ partition $I$ for all $n \geq 1$ , there exists $j\p{n,x}$ such that $x \in I_{n,j\p{n,x}}$ for all $n \geq 1$ .

Notice that this collection shrinks nicely to $x$ : $I_{n,j\p{n,x}} \subseteq B\p{x, 2^{1-n}}$ and if $m$ denotes the Lebesgue measure on $I$ ,

m\p{I_{n,j\p{n,x}}} = \frac{1}{2^{-n}} > \frac{1}{4} \cdot \frac{1}{2^{n-1}} = \frac{m\p{B\p{x, 2^{1-n}}}}{4}.

Thus, the Lebesgue differentiation theorem applies:

\begin{aligned} \abs{E_nf\p{x} - f\p{x}} &= \abs{2^n \int_{I_{n,j\p{n,x}}} f \,\diff{t} - f\p{x}} \\ &= \abs{\frac{1}{m\p{I_{n,j\p{n,x}}}} \int_{I_{n,j\p{n,x}}} f\p{t} - f\p{x} \,\diff{t}} \\ &\leq \frac{1}{m\p{I_{n,j\p{n,x}}}} \int_{I_{n,j\p{n,x}}} \abs{f\p{t} - f\p{x}} \,\diff{t}, \end{aligned}

which tends to $0$ as $n \to \infty$ , by the Lebesgue differentiation theorem. Thus, because the Lebesgue differentiation theorem applies for almost every $x$ , it follows that $E_nf\p{x} \xrightarrow{n\to\infty} f\p{x}$ almost everywhere, as required.