Differentiation — Alex Beaudin

Differentation starts with some random topics, which I’m not sure I understand why we’re studying yet.

The Vitali Covering Lemma

This is going to be useful, but for now it’s just going to be weird.

The proof is sessntially a greedy algorithm. Start with an empty set. Take the largest interval that is disjoint from your current set and add it to the set. Keep doing this until there are no more disjoint intervals. The proof then shows that $I_j$ must be subsumed in some of those intervals $* 3$ , because if not, then we can add it to the set.

Hardy-Littlewood Maximal Inequality

This function gives an upper bound-ish of the density of the function at each point.

The Markov inequality estimates the size of the set on which a function exceeds a value $c \in \R$ . The next result estimates of the size of the set on which the maximal function $h^*$ exceeds $c$ . It may seem unintuitive or ungrounded for now, but this result will play a central role in the proof of the Lebesgue Differentiation Theorem.

Proof.

Suppose $F$ is a closed bounded subset of $\{b \in \R \,:\, h^*(b) > c\}$ . We will show that $|F| \leq \frac{3}{c} \int_{-\infty}^\infty |h|$ , which implies the desired result.

By definition of $F$ , for every $b \in F$ there exists a $t_b > 0$ such that

\htmlId{eq-5}{\frac{1}{2 t_b} \int_{b - t_b}^{b + t_b} |h| > c.} \tag{5}

Clearly,

\htmlId{eq-6}{F \subseteq \bigcup_{b \in F} (b -t_b, b+t_b), } \tag{6}

and the Heine-Borel theorem tells us that this open cover of a closed bounded set must have a finite subcover. In other words, there exists $b_1, \dotsc, b_n$ such that

\htmlId{eq-7}{F \subseteq (b_1 - t_{b_1}, b_1 + t_{b_1}) \cup \cdots \cup (b_n - t_{b_n}, b_n + t_{b_n}).} \tag{7}

For convenience, let’s denote these by $I_i$ Now, by the Vitali covering lemma, we can further cut this down to disjoint intervals $I_{k_1}, \dotsc, I_{k_n}$ . It then follows that

\htmlId{eq-8}{\begin{align*} |F| &\leq |I_1 \cup \cdots \cup I_n| \\ & \leq |(3 * I_{k_1}) \cup \cdots \cup (3 * I_{k_m}) | \\ & \leq | 3 * I_{k_1} | + \cdots + |3 * I_{k_m}| \\ & = 3\left( | I_{k_1} | + \cdots + |I_{k_m}| \right) \\ & < \frac{3}{c} \left( \int_{I_{k_1}} |h| + \cdots + \int_{I_{k_m}} |h| \right) \\ & \leq \frac{3}{c} \int_{-\infty}^\infty |h|, \end{align*}} \tag{8}

where the second-to-last inequality comes from our definition of $t_b$ .

$\blacksquare$

Derivatives of Integrals

The next result is the one we’ve been waiting for. It has differntiation in its name, even though no derivative is in sight. Don’t worry, the derivatives will appear later.

The proof is long to type out. Basically, the idea is to use Luzin’s theoreom, because in the case where $f$ is continuous, the result follows trivially. We use Luzin’s theorem to bound the error, and find that it can only differ on a measure zero set.

Proof.

If $t \neq 0$ , then

\htmlId{eq-12}{\begin{align*} \left| \frac{g(b+t ) - g(b)}{t} - f(b)\right| &= \left| \frac{\int_{-\infty}^{b+t} f - \int_{-\infty}^{b}f}{t} - f(b)\right| \\ &= \left| \frac{\int_b^{b+t}f}{t} - f(b) \right| \\ &= \left| \frac{\int_b^{b+t}(f - f(b))}{t}\right| \\ &\leq \sup_{ \{x \in \R \,:\, |x - b| < |t|\} } |f(x) - f(b)|. \end{align*}} \tag{12}

If $\varepsilon > 0$ , the by the continuity of $f$ at $b$ , for small enough $t$ the last quantity is less than $\varepsilon$ . Thus, $g$ is differentiable at $b$ and $g'(b) = f(b)$ .

$\blacksquare$