Differentiation of measures, the fundamental theorems of calculus and Sobolev spaces

A measure can be obtained from another measure by integrating a function, and a function on an interval of $\mathbb{R}$ can sometimes be obtained by integrating its derivative.

Differentiation of measures#

Remark 1. Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${f \colon X \to [0, +\infty)}$ be measurable. Then ${\nu = f\mu,}$ defined by

$$\nu(E) := \int_E f \, d\mu \qquad \text{for every } E \in \mathcal{M},$$

is another measure on ${(X, \mathcal{M}),}$ by Proposition 8 of the sixth post with $f$ in place of ${\varphi.}$

The question raised by Remark 1 is the converse. Given two measures $\nu$ and $\mu$ on ${(X, \mathcal{M}),}$ does there exist $f$ such that ${\nu = f\mu?}$

Definition 2 (Absolutely continuous measure). Let ${(X, \mathcal{M})}$ be a measurable space and let ${\mu, \nu \colon \mathcal{M} \to [0, +\infty]}$ be two measures. We say that $\nu$ is absolutely continuous with respect to ${\mu,}$ and we write ${\nu \ll \mu,}$ if

$$\mu(E) = 0 \implies \nu(E) = 0 \qquad \text{for every } E \in \mathcal{M}.$$

Example 3. (i) Let ${\mu, \nu \colon \mathcal{M} \to [0, +\infty]}$ be two measures on a measurable space ${(X, \mathcal{M})}$ such that ${\nu = f\mu}$ for some measurable ${f \colon X \to [0, +\infty).}$ Then ${\nu \ll \mu{:}}$ if ${\mu(E) = 0,}$ then ${\nu(E) = \int_E f \, d\mu = 0.}$
(ii) On ${(\mathbb{R}, \mathcal{L}(\mathbb{R})),}$ consider the Lebesgue measure $\lambda$ and the Dirac delta $\delta_0$ at zero of Example 4 of the second post. Then ${\lambda \not\ll \delta_0.}$ The set ${E = [1, 2]}$ does not contain ${0,}$ so that ${\delta_0(E) = 0,}$ but ${\lambda(E) = 1 \neq 0.}$

Theorem 4 (Radon–Nikodym). Let ${(X, \mathcal{M})}$ be a measurable space and let ${\mu, \nu \colon \mathcal{M} \to [0, +\infty]}$ be two measures. Suppose that ${\nu \ll \mu}$ and that $\mu$ and $\nu$ are $\sigma$-finite. Then there exists a measurable function ${f \colon X \to [0, +\infty)}$ such that ${\nu = f\mu.}$

In Theorem 4, the measure $\mu$ is $\sigma$-finite as in Definition 8 (iii) of the second post. That is, there exists ${(E_m) \subset \mathcal{M}}$ with ${\bigcup_m E_m = X}$ and ${\mu(E_m) < +\infty}$ for every ${m.}$

A function $f$ as in Theorem 4 is called a Radon–Nikodym derivative of $\nu$ with respect to ${\mu.}$ Theorem 4 says that ${d\nu = f \, d\mu,}$ which suggests the notation ${f = d\nu/d\mu.}$

The counting measure on $[0, 1]$#

Remark 5. In Theorem 4 the assumption that $\mu$ is $\sigma$-finite is needed. Otherwise, on ${([0, 1], \mathcal{L}([0, 1])),}$ take the Lebesgue measure $\lambda$ and the counting measure $\mu$ on ${[0, 1]}$ of Example 3 of the second post. Then $\mu$ is not $\sigma$-finite, ${\lambda \ll \mu,}$ and there exists no measurable ${f \colon [0, 1] \to [0, +\infty)}$ such that ${\lambda = f\mu.}$

The measure $\mu$ is not $\sigma$-finite on ${[0, 1],}$ because ${\mu(E) < +\infty}$ if and only if $E$ is finite. If there existed ${(E_m) \subset \mathcal{L}([0, 1])}$ with ${\mu(E_m) < +\infty}$ for every $m$ and ${[0, 1] = \bigcup_m E_m,}$ then ${[0, 1]}$ would be countable, which is impossible.

Moreover ${\lambda \ll \mu.}$ If ${\mu(E) = 0,}$ then ${E = \emptyset,}$ and ${\lambda(\emptyset) = 0.}$ Finally, suppose that a measurable ${f \colon [0, 1] \to [0, +\infty)}$ satisfies ${\lambda = f\mu.}$ Then

$$1 = \lambda([0, 1]) = \int_{[0, 1]} f \, d\mu,$$

hence there exists ${E \in \mathcal{L}([0, 1])}$ such that ${\mu(E) > 0}$ and ${f > 0}$ on ${E.}$ Up to passing to a subset, ${0 < \mu(E) < +\infty,}$ hence $E$ is finite and ${\lambda(E) = 0.}$ Therefore

$$0 = \lambda(E) = \int_E f \, d\mu > 0,$$

which is impossible.

Lebesgue differentiation theorem#

Theorem 6 (Lebesgue differentiation theorem). Let ${f \in \mathcal{L}^1(\mathbb{R}^N, \mathcal{L}(\mathbb{R}^N), \lambda).}$ Then, for almost every ${x \in \mathbb{R}^N,}$

$$\lim_{r \to 0^+} \frac{1}{\lambda(B_r(x))} \int_{B_r(x)} |f(\cdot) - f(x)| \, d\lambda = 0,$$

where $B_r(x)$ is the open ball of centre $x$ and radius ${r.}$ In the integral, ${f(\cdot)}$ varies, while ${f(x)}$ is a constant. In particular, if ${N = 1,}$ then ${B_r(x) = (x - r, x + r)}$ and ${\lambda(B_r(x)) = 2r,}$ so that

$$\lim_{r \to 0^+} \frac{1}{2r} \int_{x-r}^{x+r} |f - f(x)| \, d\lambda = 0 \qquad \text{for almost every } x \in \mathbb{R}.$$

Fundamental theorems of calculus#

For continuous functions the fundamental theorem of calculus is already known.

Theorem 7 (Baby fundamental theorem of calculus). Let ${a < b}$ be real numbers, let ${f \colon [a, b] \to \mathbb{R}}$ be continuous, and define

$$F(x) = \int_a^x f(t) \, dt, \qquad x \in [a, b].$$

Then $F$ is continuous on ${[a, b],}$ and $F$ is differentiable at every ${x \in [a, b],}$ with ${F'(x) = f(x).}$ Moreover, ${F(y) - F(x) = \int_x^y f(t) \, dt}$ for all ${x, y \in [a, b]}$ with ${x < y.}$

Corollary 8. Let ${a < b}$ be real numbers and let ${g \colon [a, b] \to \mathbb{R}}$ belong to ${C^1([a, b]),}$ the set of the continuously differentiable functions on ${[a, b].}$ Then

$$g(x) = g(a) + \int_a^x g'(t) \, dt \qquad \text{for every } x \in [a, b].$$

First fundamental theorem of calculus#

The question is what can be proved, for ${f \in L^1,}$ about the function

$$F(x) = \int_a^x f \, d\lambda.$$

Theorem 9 (First fundamental theorem of calculus). Let ${-\infty < a < b < +\infty,}$ let ${f \in L^1((a, b)),}$ and let

$$F(x) := \int_a^x f \, d\lambda \qquad \text{for every } x \in [a, b].$$

Then $F$ is continuous on ${[a, b],}$ $F$ is differentiable almost everywhere on ${(a, b),}$ and ${F'(x) = f(x)}$ at almost every ${x \in (a, b).}$ Moreover, ${F(y) - F(x) = \int_x^y f \, d\lambda}$ for all ${x, y \in [a, b]}$ with ${x < y.}$

Proof (sketch). In the proof of differentiability only the case ${h > 0}$ is treated. First, let ${(x_n) \subset [a, b]}$ with ${x_n \to x_0 \in [a, b].}$ We need to prove that ${F(x_n) \to F(x_0).}$ We have

$$F(x_n) = \int_a^{x_n} f \, d\lambda = \int_a^b f \chi_{(a, x_n)} \, d\lambda,$$

where ${f \chi_{(a, x_n)} \to f \chi_{(a, x_0)}}$ almost everywhere as ${n \to \infty,}$ and ${|f \chi_{(a, x_n)}| \le |f| \in L^1((a, b))}$ for every ${n.}$ By the dominated convergence theorem, Theorem 10 of the sixth post, ${f \chi_{(a, x_n)} \to f \chi_{(a, x_0)}}$ in ${L^1((a, b)).}$ Hence

$$\begin{aligned} \lim_n F(x_n) &= \lim_n \int_a^b f \chi_{(a, x_n)} \, d\lambda \\ &= \int_a^b f \chi_{(a, x_0)} \, d\lambda = \int_a^{x_0} f \, d\lambda \\ &= F(x_0). \end{aligned}$$

Next, choose a representative ${f \in \mathcal{L}^1((a, b))}$ of the class of ${f.}$ By Theorem 6 with ${N = 1,}$ at almost every ${x \in (a, b),}$

$$\lim_{r \to 0^+} \frac{1}{2r} \int_{x-r}^{x+r} |f - f(x)| \, d\lambda = 0.$$

Now fix ${x \in (a, b)}$ such that this limit holds: such points are called Lebesgue points of ${f.}$ We want to prove that

$$\Big| \frac{F(x + h) - F(x)}{h} - f(x) \Big| \longrightarrow 0 \qquad \text{as } h \to 0.$$

For ${h > 0}$ so small that ${(x - h, x + h) \subset (a, b),}$ we compute

$$\begin{aligned} 0 \le \left| \frac{F(x + h) - F(x)}{h} - f(x) \right| &= \left| \frac{1}{h} \biggl( \int_a^{x+h} f \, d\lambda - \int_a^x f \, d\lambda \biggr) - f(x) \right| \\ &= \left| \frac{1}{h} \int_x^{x+h} f \, d\lambda - f(x) \right| \\ &= \left| \frac{1}{h} \int_x^{x+h} f \, d\lambda - \frac{1}{h} \int_x^{x+h} f(x) \, d\lambda \right| \\ &= \frac{1}{h} \left| \int_x^{x+h} \big( f - f(x) \big) \, d\lambda \right| \\ &\le \frac{1}{h} \int_x^{x+h} |f - f(x)| \, d\lambda \le \frac{2}{2h} \int_{x-h}^{x+h} |f - f(x)| \, d\lambda. \end{aligned}$$

The third equality uses that ${\alpha = (1/h) \int_x^{x+h} \alpha \, d\lambda}$ for every ${\alpha \in \mathbb{R},}$ with ${\alpha = f(x).}$ As ${h \to 0,}$ the last term tends to $0$ because $x$ is a Lebesgue point. Hence, at almost every ${x \in (a, b),}$

$$\lim_{h \to 0} \Big| \frac{F(x + h) - F(x)}{h} - f(x) \Big| = 0, \qquad \text{that is,} \qquad F'(x) = f(x). \tag*{$\blacksquare$}$$

The Cantor and Heaviside functions#

Remark 10. Let ${g \colon (a, b) \to \mathbb{R}}$ be such that $g'(x)$ exists for almost every ${x \in (a, b)}$ and ${g' \in L^1((a, b)).}$ Does it follow that “${g(x) = \int_a^x g' \, d\lambda}$”, in the sense of Corollary 8, that is, ${g(y) - g(x) = \int_x^y g' \, d\lambda}$ for ${x < y?}$ No, as (1) and (2) show.

(1) Let ${v \colon [0, 1] \to \mathbb{R}}$ be the Cantor function of Example 8 of the third post. Then $v'(x)$ exists and ${v'(x) = 0}$ for almost every ${x,}$ by Example 9 (iii) of the fourth post, and ${v' = 0 \in L^1((0, 1)).}$ But the formula fails for ${x = 0}$ and ${y = 1{:}}$

$$v(1) - v(0) = 1 \neq \int_0^1 v' \, d\lambda = \int_0^1 0 \, d\lambda = 0.$$

(2) Let ${H \colon \mathbb{R} \to \mathbb{R}}$ be the Heaviside function, equal to 0 on ${(-\infty, 0]}$ and to 1 on ${(0, +\infty),}$ as in Figure 1. Then $H'(x)$ exists and ${H'(x) = 0}$ for every ${x \neq 0,}$ but, for every ${x > 0,}$

$$\int_{-3}^x H' \, d\lambda = 0 \neq H(x) - H(-3).$$

Figure 1. The drawing of Remark 10 (2), with the point $-3$ of the formula marked on the axis: the graph of the Heaviside function ${H.}$

Absolutely continuous functions#

Definition 11 (Absolutely continuous function). Let ${a < b}$ be real numbers and ${f \colon [a, b] \to \mathbb{R}.}$ We say that $f$ is absolutely continuous on ${[a, b]}$ if for every positive $\varepsilon$ there exists a positive $\delta$ such that, for every finite family ${\{ (a_i, b_i) : i = 1, \dots, m \}}$ of pairwise disjoint subintervals of ${[a, b],}$

$$\sum_{i=1}^{m} (b_i - a_i) < \delta \qquad \text{implies} \qquad \sum_{i=1}^{m} |f(b_i) - f(a_i)| < \varepsilon.$$

Here the intervals are numbered so that ${a \le a_1 < b_1 \le a_2 < b_2 \le \dots < b_m \le b,}$ as in Figure 2, and we write ${f \in \mathrm{AC}([a, b]).}$

Figure 2. The drawing of Definition 11: the interval ${[a, b]}$ with the subintervals ${(a_1, b_1),}$ ${(a_2, b_2)}$ and so on up to ${(a_m, b_m),}$ the values ${f(a_1),}$ ${f(b_1),}$ $f(a_2)$ and $f(b_2)$ above them, and the vertical segments of lengths ${|f(b_1) - f(a_1)|}$ and ${|f(b_2) - f(a_2)|.}$

Remark 12 (Lipschitz and uniformly continuous functions). Let ${a < b}$ be real numbers and ${f \colon [a, b] \to \mathbb{R}.}$ The function $f$ is Lipschitz on ${[a, b]}$ if there exists ${L > 0}$ such that ${|f(y) - f(x)| \le L |y - x|}$ for all ${x, y \in [a, b].}$ It is uniformly continuous on ${[a, b]}$ if for every positive $\varepsilon$ there exists a positive $\delta$ such that, for all ${x < y}$ in ${[a, b],}$ ${|y - x| < \delta}$ implies ${|f(y) - f(x)| < \varepsilon.}$ In this case we write ${f \in \mathrm{UC}([a, b]).}$ Then

$$\begin{aligned} f \text{ Lipschitz on } [a, b] &\implies f \in \mathrm{AC}([a, b]) \\ &\implies f \in \mathrm{UC}([a, b]) \implies f \text{ continuous on } [a, b]. \end{aligned}$$

Lemma 13 (Absolute continuity of the integral). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${f \in L^1.}$ Then for every positive $\varepsilon$ there exists a positive $\delta$ such that, for every ${E \in \mathcal{M}}$ with ${\mu(E) < \delta,}$

$$\int_E |f| \, d\mu < \varepsilon.$$

Proof (complete). Let ${\varepsilon > 0.}$ Take a function $s$ such that (1) $s$ is simple, (2) ${0 \le s \le |f|}$ almost everywhere, and (3) ${\int_X s \, d\mu \ge \int_X |f| \, d\mu - \varepsilon/2.}$ Indeed, such an $s$ exists by Definition 3 of the fifth post, because ${\int_X |f| \, d\mu < +\infty,}$ and it can be taken with ${s \le |f|}$ at every point.

Let ${M := \max_X s.}$ We can take ${\delta := \varepsilon/(2M).}$ In detail, $M$ is a real number, because $s$ takes finitely many values and is at most $|f|$ at every point, and if ${M = 0}$ every ${\delta > 0}$ works, because (3) then gives ${\int_X |f| \, d\mu \le \varepsilon/2.}$ So take ${E \in \mathcal{M}}$ with ${\mu(E) < \delta.}$ Then

$$\begin{aligned} \int_E |f| \, d\mu &= \int_E \big( |f| - s \big) \, d\mu + \int_E s \, d\mu \\ &\le \int_X \big( |f| - s \big) \, d\mu + M \mu(E) \\ &< \frac{\varepsilon}{2} + M \frac{\varepsilon}{2M} \\ &= \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon, \end{aligned}$$

where ${|f| - s \ge 0,}$ the first inequality is monotonicity, in the domain of integration and, as ${s \le M,}$ in the function, and the second uses (3) and ${\mu(E) < \delta = \varepsilon/(2M).}$ $\blacksquare$

Second fundamental theorem of calculus#

Theorem 14 (Second fundamental theorem of calculus). Let ${a < b}$ be real numbers and let ${f \colon [a, b] \to \mathbb{R}}$ be measurable. The following are equivalent.

(1) The function $f$ is differentiable almost everywhere on ${(a, b),}$ with ${f' \in L^1((a, b)),}$ and

$$f(y) - f(x) = \int_x^y f' \, d\lambda \qquad \text{for all } x, y \in [a, b] \text{ with } x < y.$$

(2) ${f \in \mathrm{AC}([a, b]).}$
In particular, let ${g \in L^1((a, b))}$ and ${F(x) = \int_a^x g \, d\lambda}$ for ${x \in [a, b].}$ Then ${F \in \mathrm{AC}([a, b]).}$

The last claim of Theorem 14 follows from the implication from (1) to (2), since $F$ satisfies (1) by Theorem 9.

Proof (sketch). Only the implication from (1) to (2) is proved. Fix ${\varepsilon > 0}$ and let ${\delta > 0}$ be given by Lemma 13, applied on ${([a, b], \mathcal{L}([a, b]), \lambda)}$ to ${f' \in L^1.}$ This means that, for every ${E \in \mathcal{L}([a, b]),}$

$$\lambda(E) < \delta \implies \int_E |f'| \, d\lambda < \varepsilon.$$

Now we prove that ${f \in \mathrm{AC}([a, b]).}$ Let ${\{ (a_i, b_i) : i = 1, \dots, m \}}$ be pairwise disjoint subintervals of ${[a, b]}$ with total length less than ${\delta.}$ Since the intervals are pairwise disjoint,

$$\sum_{i=1}^{m} (b_i - a_i) = \sum_{i=1}^{m} \lambda((a_i, b_i)) = \lambda\Big( \bigcup_{i=1}^{m} (a_i, b_i) \Big) < \delta.$$

We compute

$$\begin{aligned} \sum_{i=1}^{m} |f(b_i) - f(a_i)| &= \sum_{i=1}^{m} \biggl| \int_{a_i}^{b_i} f' \, d\lambda \biggr| \\ &\le \sum_{i=1}^{m} \int_{(a_i, b_i)} |f'| \, d\lambda \\ &= \int_{\bigcup_{i=1}^{m} (a_i, b_i)} |f'| \, d\lambda < \varepsilon, \end{aligned}$$

where the first equality is (1), and the last inequality is the choice of ${\delta,}$ as the integral is over a set $E$ with ${\lambda(E) < \delta.}$ $\blacksquare$

Example 15. Let ${v \colon [0, 1] \to \mathbb{R}}$ be the Cantor function of Example 8 of the third post. The function $v$ is continuous on ${[0, 1],}$ but ${v \notin \mathrm{AC}([0, 1])}$ by Theorem 14. Still, ${v \in \mathrm{UC}([0, 1])}$ by the Heine–Cantor theorem: if ${u \colon [a, b] \to \mathbb{R}}$ is continuous, then ${u \in \mathrm{UC}([a, b]).}$

Example 16. There exists ${g \in \mathrm{AC}([0, 1])}$ that is not Lipschitz, for example ${g(x) = \sqrt{x}}$ for ${x \in [0, 1].}$ It satisfies ${g'(x) = 1/(2\sqrt{x})}$ for every ${x \in (0, 1)}$ and

$$g(y) - g(x) = \int_x^y \frac{1}{2\sqrt{t}} \, dt \qquad \text{for all } x, y \in [0, 1] \text{ with } x < y,$$

so that ${g \in \mathrm{AC}([0, 1])}$ by Theorem 14. In detail, ${g' \in L^1((0, 1))}$ by (2) of Theorem 17 of the fifth post, because ${g' \ge 0}$ and its generalized Riemann integral over $(0, 1)$ is ${g(1) - g(0) = 1.}$

Corollary 17 (Integration by parts in AC). Let ${a < b}$ be real numbers and ${f \in \mathrm{AC}([a, b]).}$ Then, for every ${\varphi \in C_c^\infty(a, b),}$

$$\int_a^b f \varphi' \, d\lambda = -\int_a^b f' \varphi \, d\lambda,$$

where $f'$ is the derivative of ${f,}$ defined almost everywhere.

Sobolev spaces (introduction)#

Sobolev spaces are defined through the identity of Corollary 17, with a function $w$ in place of ${f'.}$

Definition 18 (Sobolev space ${W^{1,1},}$ weak derivative). Let ${-\infty \le a < b \le +\infty}$ and ${f \in L^1((a, b)).}$ We say that ${f \in W^{1,1}((a, b)),}$ the Sobolev space $W^{1,1}$ on ${(a, b),}$ if there exists ${w \in L^1((a, b))}$ such that

$$\int_a^b f \varphi' \, d\lambda = -\int_a^b w \varphi \, d\lambda \qquad \text{for every } \varphi \in C_c^\infty(a, b).$$

The function $w$ is called a weak derivative of ${f.}$

Remark 19. In the same way one can take ${f \in L^p((a, b))}$ with a weak derivative ${w \in L^p((a, b)).}$ In this case one writes ${f \in W^{1,p}((a, b)).}$

Theorem 20. Let ${-\infty < a < b < +\infty.}$ Then ${\mathrm{AC}([a, b]) = W^{1,1}((a, b));}$ more precisely, ${f \in \mathrm{AC}([a, b])}$ if and only if ${f \in W^{1,1}((a, b)),}$ where $f$ is identified with its class in ${L^1((a, b)).}$ In particular, every ${f \in W^{1,1}((a, b))}$ has a continuous representative.

Proof (complete). First, let ${f \in \mathrm{AC}([a, b]).}$ By Theorem 14 and Corollary 17, $f'$ exists almost everywhere, ${f' \in L^1((a, b)),}$ and ${\int_a^b f \varphi' \, d\lambda = -\int_a^b f' \varphi \, d\lambda}$ for every ${\varphi \in C_c^\infty(a, b).}$ Hence ${f \in W^{1,1}((a, b)).}$

Conversely, let ${f \in W^{1,1}((a, b))}$ and let $w$ be a weak derivative of ${f.}$ Let ${F(x) = \int_a^x w \, d\lambda}$ for ${x \in [a, b].}$ Then ${F \in \mathrm{AC}([a, b])}$ by Theorem 14, and $F'$ exists and ${F' = w}$ almost everywhere by Theorem 9. Let ${\varphi \in C_c^\infty(a, b).}$ Then

$$\int_a^b F \varphi' \, d\lambda = -\int_a^b F' \varphi \, d\lambda = -\int_a^b w \varphi \, d\lambda = \int_a^b f \varphi' \, d\lambda,$$

where the first equality is Corollary 17, the second holds because ${F' = w}$ almost everywhere, and the third because ${f \in W^{1,1}((a, b)).}$ Hence

$$\int_a^b (F - f) \varphi' \, d\lambda = 0 \qquad \text{for every } \varphi \in C_c^\infty(a, b).$$

By the constancy lemma, (2) of Corollary 16 of the fifth post, there exists ${C \in \mathbb{R}}$ such that ${F - f = C}$ almost everywhere. Hence ${f = F - C}$ almost everywhere, where ${F \in \mathrm{AC}([a, b])}$ and $C$ is a constant, so $f$ is equal almost everywhere to a function of ${\mathrm{AC}([a, b]).}$ Therefore $f$ has a representative in ${\mathrm{AC}([a, b]).}$ $\blacksquare$