Differentiation

Average Value over an Interval

Suppose that

f \in L^{1} (ℝ)

and

I \subseteq ℝ

is an interval with positive finite Lebesgue measure. The average value of

f

over

I

\frac{1}{λ (I)} \int_{I} f d λ .

Lebesgue Point

Suppose that

f \in L_{loc}^{1} (ℝ)

. A point

a \in ℝ

is a Lebesgue point of

f

\lim_{r ↓ 0} \frac{1}{2 r} \int_{a - r}^{a + r} | f (x) - f (a) | d λ (x) = 0.

Lebesgue Differentiation Theorem

Suppose that

f \in L_{loc}^{1} (ℝ)

. Then almost every

a \in ℝ

is a Lebesgue point of

f

It is enough to prove the claim for $f \in L^{1} (ℝ)$ , then apply it on bounded intervals for the local case. Approximate $f$ in $L^{1}$ by a continuous compactly supported function $g$ , using approximation in $L^{1}$ by continuous functions. Every point is a Lebesgue point of $g$ by continuity. The bad set for $f$ is controlled by the Hardy-Littlewood maximal function of $f - g$ ; the maximal inequality makes its measure arbitrarily small as ${‖ f - g ‖}_{1} \to 0$ . Hence the non-Lebesgue points of $f$ form a null set.

Differentiation of Indefinite Integrals

Suppose that

f \in L^{1} (ℝ)

and

F (x) : = \int_{- \infty}^{x} f d λ .

Then

F^{'} (x) = f (x)

for almost every

x \in ℝ

If $a$ is a Lebesgue point of $f$ , then $\frac{F (a + h) - F (a)}{h} - f (a) = \frac{1}{h} \int_{a}^{a + h} (f - f (a)) d λ$ for $h > 0$ , with the analogous formula for $h < 0$ . The absolute value of the right side is bounded by the average of $| f - f (a) |$ over an interval shrinking to $a$ , which tends to $0$ . By the Lebesgue differentiation theorem, this holds for almost every $a$ .

Hardy-Littlewood Maximal Function

Suppose that

f \in L_{loc}^{1} (ℝ)

. The Hardy-Littlewood maximal function of

f

M f (x) : = \sup_{r \in ℝ^{+}} \frac{1}{2 r} \int_{x - r}^{x + r} | f | d λ .

Hardy-Littlewood Maximal Inequality

Suppose that

f \in L^{1} (ℝ)

and

α \in ℝ^{+}

. Then

λ ({x \in ℝ : M f (x) > α}) \leq \frac{3}{α} {‖ f ‖}_{1} .

For each $x$ with $M f (x) > α$ , choose an interval $I_{x}$ centered at $x$ such that $\int_{I_{x}} | f | d λ > α λ (I_{x})$ . On any bounded subcollection, the one-dimensional covering lemma selects disjoint intervals $I_{j}$ whose triples cover the same centers. Therefore $λ ({M f > α} \cap [- N, N]) \leq 3 \sum_{j} λ (I_{j}) \leq \frac{3}{α} \int | f | d λ .$ Letting $N \to \infty$ and using the measure of an increasing union proves the inequality.

Absolutely Continuous Function on an Interval

Suppose that

a < b

. A function

F : [a, b] \to ℝ

is absolutely continuous if for every

ϵ \in ℝ^{+}

, there exists

δ \in ℝ^{+}

such that for every finite pairwise disjoint collection of intervals

(a_{k}, b_{k}) \subseteq [a, b]

\sum_{k} (b_{k} - a_{k}) < δ ⟹ \sum_{k} | F (b_{k}) - F (a_{k}) | < ϵ .

Absolutely Continuous Functions are Integrals of Their Derivatives

Suppose that

F : [a, b] \to ℝ

is absolutely continuous. Then

F^{'}

exists almost everywhere,

F^{'} \in L^{1} ([a, b])

, and

F (y) - F (x) = \int_{x}^{y} F^{'} d λ

whenever

a \leq x \leq y \leq b

Absolute continuity implies bounded variation, so $F$ is differentiable almost everywhere and $F^{'} \in L^{1} ([a, b])$ . Define $G (x) = F (a) + \int_{a}^{x} F^{'} d λ .$ By differentiation of indefinite integrals, $G^{'} = F^{'}$ almost everywhere. Thus $H = F - G$ is absolutely continuous with derivative $0$ almost everywhere. Applying the absolute-continuity condition to the open set where $H$ changes by more than a fixed amount shows $H$ is constant. Since $H (a) = 0$ , $H = 0$ , giving the displayed integral formula.