<Home>Qualifying Exams><Analysis>Spring 2016 - Problem 3

Spring 2016 - Problem 3

Lebesgue differentiation theorem

Let $f \in L^1_{\mathrm{loc}}\p{\R}$ be real-valued and assume that for each integer $n > 0$ , we have

f\p{x + \frac{1}{n}} \geq f\p{x},

for almost all $x \in \R$ . Show that for each real number $a \geq 0$ , we have

f\p{x + a} \geq f\p{x}

for almost all $x \in \R$ .

Solution.

Let $c < d$ and observe that for any $n, k \in \N$ ,

\int_c^d f\p{y + \frac{k}{n}} - f\p{y} \,\diff{y} \geq \int_c^d f\p{y + \frac{1}{n}} - f\p{y} \,\diff{y} \geq 0

by iterating the inequality. This shows that the inequality holds for any non-negative rational. Let $a \geq 0$ and let $\set{q_n}_n \subseteq \Q$ be a sequence which converges to $a$ . Since translation is continuous on $L^1\p{\br{c,d}}$ , we get

\int_c^d f\p{y + q_n} - f\p{y} \,\diff{y} \xrightarrow{n\to\infty} \int_c^d f\p{y + a} - f\p{y} \,\diff{y} \geq 0,

since each term in the sequence was non-negative. Almost every $x \in \R$ is a Lebesgue point, so it suffices to show that the claim is true for $x$ a Lebesgue point of both $f\p{y}$ and $f\p{y + a}$ . Since $c, d$ were arbitrary in the above, we may set $\br{c, d} = B\p{x, r}$ . Hence,

\int_{x-r}^{x+r} f\p{y + a} - f\p{y} \,\diff{y} \geq 0 \implies \frac{1}{2r} \int_{x-r}^{x+r} f\p{y + a} - f\p{y} \,\diff{y} \geq 0,

so by the Lebesgue differentiation theorem, the average tends to

f\p{x + a} - f\p{x} \geq 0,

which completes the proof.