Differentiation of the Lebesgue integral. Absolutely continuous functions

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Unlike the previous chapters, here throughout \(X = \mathbb {R}\), and the measure is Lebesgue measure \(\lambda\) (or a Lebesgue-Stieltjes measure generated by it), unless explicitly stated otherwise. We apply the abstract theory of measure (in particular, the Radon-Nikodym theorem from the chapter on signed measures) to this concrete case and obtain classical facts of mathematical analysis — differentiation of the integral, the Newton-Leibniz formula, and the relationship between a distribution’s density and the derivative of its distribution function.

Throughout the first three sections, the measure is Lebesgue measure \(\lambda\) on \(\mathbb {R}\). The notation \(f \in L^1(\mathbb {R})\) means that \(f: \mathbb {R} \to \mathbb {R}\) is Lebesgue measurable and \(\left\| f\right\|_{1} = \int_{\mathbb {R}} \left|f\right| \, d\lambda < \infty\). For integrals over segments we use the abbreviation

\[ \int _{a}^{b} f := \int _{[a,b]} f \, d\lambda \]

1 The Hardy-Littlewood maximal function

Definition 1 Let \(I\) be a bounded nonempty open interval in \(\mathbb {R}\). Denote by \(3 * I\) the open interval with the same center as \(I\) and three times the length.

For example, if \(I = (0, 10)\), then \(3 * I = (-10, 20)\).

Lemma 1 (Vitali covering lemma, finite version) Let \(I_1, \ldots , I_n\) be a list of bounded nonempty open intervals in \(\mathbb {R}\). Then there exists a disjoint sublist \(I_{k_1}, \ldots , I_{k_m}\) such that \[ I_1 \cup \ldots \cup I_n \subseteq (3 * I_{k_1}) \cup \ldots \cup (3 * I_{k_m}) \]

Example: let \(n = 4\) and

\[ I_1 = (0, 10), \quad I_2 = (9, 15), \quad I_3 = (14, 22), \quad I_4 = (21, 31) \]

Then

\[ 3*I_1 = (-10, 20), \quad 3*I_2 = (3,21), \quad 3*I_3 = (6, 30), \quad 3*I_4 = (11, 41) \]

and \(I_1 \cup I_2 \cup I_3 \cup I_4 \subseteq (3*I_1) \cup (3*I_4)\), i.e. the pair \(I_1, I_4\) works as a sublist (as is not hard to check, no other sublist works in this example).

ProblemProblem 1

Prove the step left in the proof of the Vitali lemma as an exercise: if \(I, J\) are bounded nonempty intervals, \(I \cap J \neq \varnothing\), and \(\left|J\right| \geq \left|I\right|\), then \(I \subseteq 3 * J\).

ProblemProblem 2

Show that the constant \(3\) in the Vitali lemma cannot be replaced by any smaller one.

HintHint

It suffices to consider a suitable pair of overlapping intervals of nearly equal length.

Definition 2 Let \(h: \mathbb {R} \to \mathbb {R}\) be Lebesgue measurable. The Hardy-Littlewood maximal function is the function \(h^*: \mathbb {R} \to [0, +\infty ]\) defined by: \[ h^*(b) := \sup _{t > 0} \; \frac{1}{2t}\int _{b-t}^{b+t} \left|h\right| \] In other words, \(h^*(b)\) is the supremum of the average values of \(\left|h\right|\) over all intervals centered at \(b\).

ExampleExample 1

Find \(\left(\; \mathbb {1}_{[0,1]}\right)^*\).

SolutionSolution

Let \(h = \; \mathbb {1}_{[0,1]}\). The average value of \(\left|h\right|\) over the interval \((b-t, b+t)\) equals \(\frac{\lambda \left((b-t,\, b+t) \cap [0,1]\right)}{2t} \leq 1\).

Let \(b \leq 0\). As long as \(b + t \leq 1\), the average equals \(\frac{\max (b+t,\, 0)}{2t}\); as a function of \(t\) it is nondecreasing (for \(b+t > 0\) the derivative with respect to \(t\) equals \(\frac{-2b}{(2t)^2} \geq 0\)), i.e. it is advantageous to increase \(t\) up to \(t = 1 - b\). For \(t > 1-b\) the numerator is fixed at \(1\), and the average \(\frac{1}{2t}\) decreases. The maximum is attained at \(t = 1-b\):

\[ h^*(b) = \frac{1}{2(1-b)} \]

For \(0 < b < 1\), sufficiently small intervals around \(b\) lie entirely in \([0,1]\), the average over them equals \(1\), and the average cannot exceed \(1\): \(h^*(b) = 1\). The case \(b \geq 1\) is symmetric to the case \(b \leq 0\): \(h^*(b) = \frac{1}{2b}\).

AnswerAnswer

\(\left(\; \mathbb {1}_{[0,1]}\right)^*(b) = \begin{cases} \frac{1}{2(1-b)}, & b \leq 0 \\ 1, & 0 < b < 1 \\ \frac{1}{2b}, & b \geq 1\end{cases}\)

Note: in this example \(h \in L^1(\mathbb {R})\), but \(h^*\) decays at infinity only like \(\frac{1}{2\left|b\right|}\), i.e. \(h^* \not\in L^1(\mathbb {R})\). The maximal function of an integrable function need not itself be integrable! This makes the following inequality all the more valuable: it controls the “tails” of \(h^*\) — in exactly the same spirit in which Markov’s inequality controls the tails of \(\left|h\right|\).

ProblemProblem 3

Find a formula for the Hardy-Littlewood maximal function of \(\; \mathbb {1}_{[0,1] \cup [2,3]}\).

Theorem 1 (The Hardy-Littlewood maximal inequality) Let \(h \in L^1(\mathbb {R})\). Then for any \(c > 0\) \[ \lambda \left(\left\{ b \in \mathbb {R} \; : \; h^*(b) > c\right\} \right) \leq \frac{3}{c} \left\| h\right\| _{1} \]

Compare this with Markov’s inequality (theorem Markov’s inequality): the structure is the same, but the constant has worsened from \(1\) to \(3\) — this is the price paid for the supremum over all intervals. At the same time, the maximal inequality is substantially deeper than Markov’s inequality.

ProblemProblem 4

Let \(h: \mathbb {R} \to \mathbb {R}\) be Lebesgue measurable. Prove that the set \(\left\{ b \in \mathbb {R} \; : \; h^*(b) > c\right\}\) is open for any \(c \in \mathbb {R}\). Deduce from this that \(h^*\) is Borel measurable.

ProblemProblem 5

Prove or disprove: if \(h: \mathbb {R} \to [0, +\infty )\) is a nondecreasing function, then \(h^*\) is a nondecreasing function.

ProblemProblem 6

Let \(h \in L^1(\mathbb {R})\). Prove that \(\lambda \left(\left\{ b \in \mathbb {R} \; : \; h^*(b) = \infty \right\} \right) = 0\).

ProblemProblem 7

Let \(h \in L^1(\mathbb {R})\). Prove the following strengthening of the maximal inequality: \[ \lambda \left(\left\{ b \in \mathbb {R} \; : \; h^*(b) \geq c\right\} \right) \leq \frac{3}{c}\left\| h\right\| _{1}, \qquad \forall c > 0 \]

A much deeper strengthening is a reduction of the constant \(3\) itself. The question of the best constant remained open until 2003, when A. Melas proved that the optimal constant equals \(\frac{11 + \sqrt{61}}{12} \approx 1.567\) (Annals of Mathematics 157 (2003), 647–688).

HintHint

Apply the inequality with strict “\(>\)” to the threshold \(c - \varepsilon\) and let \(\varepsilon \downarrow 0\).

2 The Lebesgue differentiation theorem

The following theorem states: for an integrable function, the average deviation from its own value is small at almost every point, when averaging over small intervals. No derivatives are yet visible in the statement, but this is exactly where all the substantive work happens — the version with derivatives will follow from it almost for free.

We will need one approximation fact, which we accept without proof: continuous functions are dense in \(L^1(\mathbb {R})\).

Lemma 2 (Without proof) Let \(f \in L^1(\mathbb {R})\), \(\varepsilon > 0\). Then there exists a continuous function \(g: \mathbb {R} \to \mathbb {R}\) such that \(\left\| f - g\right\|_{1} < \varepsilon\).

Idea of the proof: by the construction of the integral, \(f\) is approximated in \(L^1\) by simple functions; the indicator of a measurable set of finite measure is approximated by the indicator of a finite union of intervals (this follows from the construction of Lebesgue measure via covers); finally, a step function is turned into a continuous one by rounding off the corners.

Theorem 2 (Lebesgue differentiation theorem, first version) Let \(f \in L^1(\mathbb {R})\). Then \[ \lim _{t \downarrow 0} \frac{1}{2t}\int _{b-t}^{b+t} \left|f - f(b)\right| = 0 \] for almost all \(b \in \mathbb {R}\).

Before reading the proof, let us understand why the statement is obvious at every point of continuity of \(f\):

\[ \frac{1}{2t}\int _{b-t}^{b+t}\left|f - f(b)\right| \leq \sup \left\{ \left|f(x) - f(b)\right| \; : \; \left|x - b\right| \leq t\right\} \xrightarrow [t \downarrow 0]{} 0 \]

All the work consists in transferring this property from continuous functions to arbitrary integrable ones, at the cost of a set of measure zero. The scheme “approximate by a continuous function + control the approximation error with the maximal inequality” is a standard technique used to prove many “almost everywhere” theorems in analysis.

Let us turn to derivatives. Recall that the derivative of a function \(g\) at a point \(b\) is

\[ g'(b) = \lim _{t \to 0}\frac{g(b+t) - g(b)}{t}, \]

if this limit exists.

Theorem 3 (Lebesgue’s theorem on differentiation of the integral) Let \(f \in L^1_{loc}(\mathbb {R})\) be locally integrable (of arbitrary sign), let \(c\in \mathbb {R}\) be fixed, and let \(g(x) := \int_{[c,x]} f \, d\lambda\). Then \(g\) is differentiable almost everywhere and \(g'(b) = f(b)\) for almost all \(b \in \mathbb {R}\). Moreover, at every point of continuity of \(f\) this holds without exception — this is exactly the classical Newton-Leibniz formula.

A function in \(L^1_{loc}(\mathbb {R})\) may fail to be continuous at any point whatsoever — and even “incurably” so: in such a way that no modification on a set of measure zero adds even a single point of continuity (we shall construct such an example at the end of the section). For such a function, the classical Newton-Leibniz formula says nothing at all. However, as the theorem asserts, almost everywhere everything is fine, with no continuity whatsoever needed.

We can now answer the question posed at the beginning of the section.

ExampleExample 2

Does there exist a Lebesgue-measurable set \(E \subseteq [0,1]\) such that \[ \lambda \left(E \cap [0,b]\right) = \frac{b}{2} \qquad \forall b \in [0,1]? \]

SolutionSolution

Suppose such an \(E\) existed. Set \[ g(b) := \int _{-\infty }^{b} \; \mathbb {1}_{E} \] Then \(g(b) = \lambda \left(E \cap [0,b]\right) = \frac{b}{2}\) for \(b \in [0,1]\), hence \(g'(b) = \frac{1}{2}\) for all \(b \in (0,1)\). But by Lebesgue’s theorem on differentiation of the integral (Theorem 3), \(g'(b) = \; \mathbb {1}_{E}(b)\) for almost all \(b\), while the indicator never takes the value \(\frac{1}{2}\). We obtain a contradiction on a set of positive measure \((0,1)\). Hence no such \(E\) exists.

AnswerAnswer

No such set exists.

Corollary 1 (An integrable function equals its local average a.e.) Let \(f \in L^1(\mathbb {R})\). Then \[ f(b) = \lim _{t \downarrow 0}\frac{1}{2t}\int _{b-t}^{b+t} f \] for almost all \(b \in \mathbb {R}\).

Mn-proofProof

We have \[ \left|\frac{1}{2t}\int _{b-t}^{b+t} f \; - \; f(b)\right| = \left|\frac{1}{2t}\int _{b-t}^{b+t}\left(f - f(b)\right)\right| \leq \frac{1}{2t}\int _{b-t}^{b+t}\left|f - f(b)\right|, \] and the right-hand side tends to zero almost everywhere by theorem Theorem 2.

The remarkable feature of this statement is that \(f\) may be discontinuous everywhere — and yet at almost every point it is recovered from its local averages. Note that it is precisely two-sided averages (over intervals centered at \(b\)) that generalize correctly to \(\mathbb {R}^{n}\): there one averages over balls centered at the point.

ProblemProblem 8

Let \(f \in L^1(\mathbb {R})\). Prove that \(\left|f(b)\right| \leq f^*(b)\) for almost all \(b \in \mathbb {R}\).

HintHint

Use corollary Corollary 1.

ProblemProblem 9

Let \(h \in L^1(\mathbb {R})\) and \(\int_{-\infty }^{s} h = 0\) for all \(s \in \mathbb {R}\). Prove that \(h = 0\) almost everywhere.

This is a uniqueness statement: an integrable function is uniquely (up to a set of measure zero) recovered from its “accumulated integrals.” In particular, a distribution’s density is unique up to a.e. — compare with the uniqueness of the Radon-Nikodym derivative in the chapter on signed measures.

ProblemProblem 10

Let \(f: \mathbb {R} \to \mathbb {R}\) be Lebesgue measurable and \(f^2 \in L^1(\mathbb {R})\). Prove that \[ \lim _{t \downarrow 0} \frac{1}{2t}\int _{b-t}^{b+t}\left|f - f(b)\right|^2 = 0 \] for almost all \(b \in \mathbb {R}\).

HintHint

\(f\) itself may fail to lie in \(L^1(\mathbb {R})\), but \(f \cdot \; \mathbb {1}_{[-k,k]} \in L^1(\mathbb {R})\) for each \(k\) (check this via the Cauchy-Bunyakovsky inequality). Expand the square and apply corollary Corollary 1 to \(f^2 \; \mathbb {1}_{[-k,k]}\) and to \(f\; \mathbb {1}_{[-k,k]}\).

3 Absolutely continuous functions and measures

Up to now we have differentiated the integral of a specific integrable function. Now let us connect this with the general theory of signed measures (the chapter on signed measures): consider an arbitrary Lebesgue-Stieltjes measure \(\mu = \mu_F\) on \(\mathbb {R}\) and ask when it has a density with respect to \(\lambda\) — and how this density is related to the ordinary, high-school derivative of the function \(F\).

3.1 A Lebesgue-Stieltjes measure is a locally finite measure

Recall that a Lebesgue-Stieltjes measure is defined for an arbitrary nondecreasing right-continuous function \(F: \mathbb {R} \to \mathbb {R}\) via \(\mu_{F}((a,b]) = F(b) - F(a)\), \(-\infty < a < b < +\infty\). Note that a Lebesgue-Stieltjes measure is sigma-finite:

\[ \mathbb {R} = \bigcup _{n=1}^{+\infty }(-n,n], \quad \text{ with } \quad \mu _{F}((-n,n]) = F(n) - F(-n) < +\infty \]

Moreover, a Lebesgue-Stieltjes measure is finite on every bounded interval \((a,b)\). This property is actually stronger than sigma-finiteness on \(\mathbb {R}\) and is equivalent to every point \(x \in \mathbb {R}\) having a neighborhood of finite measure (this is the so-called local finiteness).

ProblemProblem 11

Let \(\mu\) be a (nonnegative) measure on \((\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))\).

1. Prove that \(\mu\) is finite on every bounded interval \(\iff\) \(\mu\) is locally finite;

2. Prove that local finiteness implies \(\sigma\)-finiteness of \(\mu\);

3. Show that the converse of the previous part fails: give an example of a (nonnegative) \(\sigma\)-finite measure \(\mu\) that is not locally finite.

So, every Lebesgue-Stieltjes measure is finite on bounded intervals. Is the converse true? Yes! It is not hard to show that if \(\mu\) is a (nonnegative) measure, finite on bounded intervals, then setting

\[ F(x) := \begin{cases} \mu ((0, x]), & 0 < x \\ -\mu ((x, 0]), & x \leq 0 \end{cases} \]

we obtain a nondecreasing right-continuous finite function, with \(\mu = \mu_F\). Let us state this result.

Lemma 3 Let \(\mu\) be a (nonnegative) measure on \((\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))\). The following conditions are equivalent:

1. \(\mu\) is a Lebesgue-Stieltjes measure (for some nondecreasing right-continuous \(F: \mathbb {R} \to \mathbb {R}\))

2. \(\mu\) is finite on bounded intervals.

3. \(\mu\) is locally finite.

Note that if \(\mu\) is a finite measure (for example, a probability measure), then instead of the \(F\) defined above one may take

\[ F(x) = \mu ((-\infty , x]) \]

i.e. the distribution function.

3.2 An absolutely continuous Lebesgue-Stieltjes measure

Let us now consider not just a Lebesgue-Stieltjes measure \(\mu = \mu_F\), but one that is absolutely continuous (with respect to Lebesgue measure): \(\mu \ll \lambda\). We know (the chapter on signed measures) that \(\mu_F\) is completely characterized by the function \(F\) — so the condition \(\mu_F \ll \lambda\) must somehow transfer to \(F\). Let us start with the cheapest observation: \(F\) will be continuous. In fact, a stronger statement holds: \(F\) will be locally uniformly continuous, i.e. uniformly continuous on every segment \([a,b]\).

ExampleExample 3

Show that if \(\mu_F \ll \lambda\), then \(F\) is locally uniformly continuous.

SolutionSolution

Indeed, on \([a,b]\) the measure \(\mu_F\) is finite, so by the lemma on the equivalent condition for absolute continuity of a measure, if \(\varepsilon > 0\) is an arbitrary fixed number, there exists \(\delta > 0\) such that for all \(a \leq x < y \leq b\) with \(\left|y - x\right| = \lambda ([x, y]) < \delta\) we have \[ \left|F(y) - F(x)\right| = F(y) - F(x) = \mu _F([x,y]) < \varepsilon \]

We have obtained a fairly strong restriction on \(F\) — local uniform continuity. That is good. However, is it enough for \(F\) to be locally uniformly continuous for \(\mu_F\) to have a density? Actually, no!

ExampleExample 4

Give an example of a nondecreasing, right-continuous, locally uniformly continuous function \(F: \mathbb {R}\to \mathbb {R}\) such that \(\mu_F\) is not absolutely continuous, i.e. does not have a density with respect to Lebesgue measure.

SolutionSolution

Let \(F\) be the Cantor staircase. Outside the segment \([0,1]\) it is constant. Hence it is uniformly continuous on any segment outside \([0,1]\).

What about on \([0,1]\) (or on any subsegment of \([0,1]\))? It is continuous there, hence it is uniformly continuous, since \([0,1]\) is compact.

In summary, the Cantor staircase \(F\) is locally uniformly continuous.

But is \(\mu_F\) absolutely continuous? No! Indeed, consider the Cantor set \(C\). \(\lambda (C) = 0\), \(\mu_F(C) = 1\). This violates absolute continuity.

We need a stronger form of continuity on a segment than uniform continuity.

3.3 Absolutely continuous functions

Note that in the uniform continuity condition on \([a,b]\) above, one could take not just \(2\) points \(x,y\), but arbitrarily many consecutive pairs of such points. Namely, for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that for every \(N \in \mathbb {N}\) and for all

\[ a \leq x_1 < y_1 \leq x_2 < y_2 \leq \ldots \leq x_{N} < y_{N} \leq b, \qquad \text{ such that } \qquad \sum _{k=1}^{N}(y_k - x_k) < \delta \]

we have

\[ \sum _{k=1}^N \left|F(y_k) - F(x_k)\right| < \varepsilon \]

A function satisfying such conditions is called absolutely continuous on the segment \([a,b]\); locally absolutely continuous on \(\mathbb {R}\) if it is absolutely continuous on every segment. In other words, an absolutely continuous function does not oscillate on a set of measure zero.

Absolute continuity is stronger than uniform continuity on a segment \([a,b]\). And strictly stronger.

ProblemProblem 12

Show that the Cantor staircase is uniformly continuous but is not an absolutely continuous function on \([0,1]\).

We now have 2 notions of absolute continuity: for measures and for functions. The following theorem shows that for Lebesgue-Stieltjes measures these are exactly one and the same notion — and, moreover, connects it with the Radon-Nikodym derivative and with the ordinary, high-school derivative.

3.4 Synthesis: the Radon-Nikodym derivative is the ordinary derivative

We already know three things separately: the abstract Radon-Nikodym theorem (theorem Radon-Nikodym from the chapter on signed measures), the Lebesgue differentiation theorem (earlier in this chapter), and the notion of an absolutely continuous function (just introduced). Let us bring them together.

Theorem 4 (The “fundamental theorem” of Lebesgue integration on \(\mathbb {R}\)) Let \(\mu = \mu_F\) be a Lebesgue-Stieltjes measure on \((\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))\). The following conditions are equivalent:

1. \(\mu \ll \lambda\);

2. \(F\) is locally absolutely continuous;

3. there exists a locally integrable \(f: \mathbb {R} \to [0,+\infty ]\) such that \(F(x) = F(c) + \int_{[c,x]} f \, d\lambda\) for all \(c, x \in \mathbb {R}\) (in short: \(d\mu = f \, d\lambda\)). Moreover, in this case \(f = F'\) \(\lambda\)-a.e. (and such an \(f\) is unique up to agreement a.e.), i.e. \(\mu (E) = \int_E F' \, d\lambda\) for all Borel \(E\).

Mn-remarkRemark

3.5 the case of non-monotone \(F\)

If \(F\) is not necessarily monotone (for example, \(F = F_1 - F_2\) for some nondecreasing \(F_1, F_2\) — say, a function of bounded variation), no new theorem needs to be proved: the corresponding real Lebesgue-Stieltjes measure \(\nu := \mu_{F_1} - \mu_{F_2}\) is decomposed via Jordan decomposition, theorem Theorem 4 is applied to each of the (nonnegative) parts separately, and the results are added — this is exactly the same linearity we already used in the chapter on signed measures to pass from nonnegative measures to real ones, and at the very beginning of this chapter to pass from \(f \geq 0\) to an arbitrary integrable \(f = f^+ - f^-\).

3.6 Practice

So, where have we arrived? What is the practical meaning of this theory? We now have, for example, the ability to determine from the distribution function whether a random variable is absolutely continuous. Moreover, we can find the density: it turns out that it suffices simply to differentiate the distribution function.

ExampleExample 5

Let a random variable \(X\) have the following distribution function (here and below \(c: [0,1] \to [0,1]\) is the Cantor staircase, i.e. the Cantor function): \[ F(x)= \begin{cases} \frac{2}{10}e^{x+2}, & -\infty < x< -2 \\ \frac{1}{10}c(x+2) + \frac{3}{10}, & -2 \leq x < -1 \\ \frac{1}{10}x^3 + \frac{6}{10}, & -1 \leq x < \sqrt[3]{2} \\ 0.85, & \sqrt[3]{2} \leq x < 5 \\ 1 - \frac{1}{2x}, & 5 \leq x < +\infty \end{cases} \]

Denote by \(\mu_F = \mathbf{P}_X\) the distribution of this random variable.

1. Decompose \(\mu_F\) into an absolutely continuous part \(\mu_a\) and a singular part \(\mu_s\) with respect to Lebesgue measure;

2. For the absolutely continuous part \(\mu_a\) find the density (Radon-Nikodym derivative) \(\frac{d\mu_a}{d\lambda }\);

3. Is the singular part \(\mu_s\) (relative to Lebesgue measure) absolutely continuous with respect to counting measure \(\#_{\mathbb {N}}\) on the natural numbers? What about with respect to \(\#_{\mathbb {Z}}\) on the integers? If so, find the Radon-Nikodym derivative.

4. Find the expectation \(\mathbb {E}\left[\sqrt{\left|X+2\right|}\right]\).

SolutionSolution

Enum-item(1)

We need to extract the locally absolutely continuous component from the function \(F\). We remove all singular fragments from \(F\). Note that the function \(F_a\), defined below, is continuous and piecewise continuously differentiable on \(\mathbb {R}\), hence locally absolutely continuous.

\[ F_a(x)= \begin{cases} \frac{2}{10}e^{x+2}, & -\infty < x< -2 \\ 0.2, & -2 \leq x < -1 \\ \frac{1}{10}x^3 + \frac{3}{10}, & -1 \leq x < \sqrt[3]{2} \\ 0.5, & \sqrt[3]{2} \leq x < 5 \\ 0.6 - \frac{1}{2x}, & 5 \leq x < +\infty \end{cases} \]

Moreover \(F_{s} := F - F_{a}\) looks as follows:

\[ F_{s}(x)= \begin{cases} 0, & -\infty < x< -2 \\ \frac{1}{10}c(x+2) + \frac{1}{10}, & -2 \leq x < -1 \\ 0.3, & -1 \leq x < \sqrt[3]{2} \\ 0.35, & \sqrt[3]{2} \leq x < 5 \\ 0.4, & 5 \leq x < +\infty \end{cases} \]

Note that \(F_s\) generates a Lebesgue-Stieltjes measure whose support is the set \(\left\{ -2, 1, \sqrt[3]{2}, 5\right\} \cup (C-2)\), where \(C-2\) is the Cantor set shifted right by \(2\). This support has Lebesgue measure \(0\). Hence this measure is singular with respect to Lebesgue measure.

In summary, the absolutely continuous component is the Lebesgue-Stieltjes measure generated by \(F_{a}\) (\(\mu_{a} := \mu_{F_{a}}\)), and the singular component is \(\mu_{s} := \mu_{F_{s}}\).

Enum-item(2)

\(F_{a}\) is absolutely continuous, so by theorem Theorem 4 it is differentiable a.e., and its derivative is exactly the density (with respect to Lebesgue measure). We have

\[ \frac{d\mu _a}{d\lambda }(x) = F'_a(x)= \begin{cases} \frac{2}{10}e^{x+2}, & -\infty < x< -2 \\ 0, & -2 < x < -1 \\ \frac{3}{10}x^2, & -1 < x < \sqrt[3]{2} \\ 0, & \sqrt[3]{2} < x < 5 \\ \frac{1}{2x^2}, & 5 < x < +\infty \\ \text{undefined}, & x \in \left\{ -2, -1, \sqrt[3]{2}, 5\right\} \end{cases} \]

At the points \(\left\{ -2, -1, \sqrt[3]{2}, 5\right\}\) one may define \(f_a(x) = \frac{d\mu_a}{d\lambda }(x)\) however one likes. For example, by zeros.

Enum-item(3)

Consider the set \(A = \left(C-2\right) \; \cap \; (-2, -1)\), i.e. the Cantor set shifted \(2\) to the left, without the points \(-2, -1\). Note that \(\mu_{s}(A) = 0.1\), while \(\#_{\mathbb {N}}(A) = 0\) and \(\#_{\mathbb {Z}}(A) = 0\). Hence \(\mu_{s}\) is not absolutely continuous with respect to counting measure on \(\mathbb {Z}\) or on \(\mathbb {N}\).

Enum-item(4)

By definition

\[ \mathbb {E}\left[\sqrt{\left|X+2\right|}\right] = \int _{\Omega } \sqrt{\left|X(\omega ) + 2\right|} \; d\mathbb {P}(\omega ) \]

Let us use a change of variables in the Lebesgue integral: \(X(\omega ) = x\). Then \(\omega = X^{-1}(x)\), \(\mathbb {P}(\omega ) = \mathbb {P}(X^{-1}(x)) = \mathbf{P}_{X}(x)\). Hence,

\[ \mathbb {E}\left[\sqrt{\left|X+2\right|}\right] = \int _{\mathbb {R}} \sqrt{\left|x+2\right|} \; d\mathbf{P}_{X}(x) \]

We decomposed \(\mathbf{P}_{X}\) as the sum \(\mu_{a} + \mu_{s}\). Let us find the integrals over each measure separately.

• \(\mu_{a}\) has density \(f_a\) with respect to Lebesgue measure. Hence,

  \[ \begin{align} \int _{\mathbb {R}} \sqrt{\left|x + 2\right|} \; d\mu _{a} & = \int _{\mathbb {R}} \sqrt{\left|x + 2\right|} \cdot f_a(x) \; d\lambda =\\ & = \int _{-\infty }^{-2}\sqrt{\left|x+2\right|} \cdot \frac{2}{10}e^{x+2} \; dx + \int _{-1}^{\sqrt[3]{2}}\sqrt{x+2} \cdot \frac{3}{10}x^2 \; dx + \int _{5}^{+\infty } \sqrt{x+2} \cdot \frac{1}{2x^2} \; dx = \\ & = \frac{2}{10}\Gamma (3/2) +\left[ \frac{3}{35}u^{7/2} -\frac{12}{25}u^{5/2} +\frac{4}{5}u^{3/2} \right]_{u=1}^{\, u=2+2^{1/3}} +\frac{\sqrt{7}}{10} -\frac{1}{4\sqrt{2}} \ln \! \left(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}\right). \approx \\ & \approx 1.107 \end{align} \]

  We omit the detailed computation of the Riemann integrals, since that is not the point of the example; what matters is understanding the general principle.

• How do we find the integral with respect to \(\mu_{s}\)? Note that

  \[ \mu _{s} = 0.1 \cdot \delta _{-2} + 0.1\cdot \mu _{C-2} + 0.1 \cdot \delta _{-1} + 0.05 \cdot \delta _{\sqrt[3]{2}} + 0.05 \cdot \delta _{5} \]

  where \(\mu_{C-2}\) is the Cantor measure shifted \(2\) to the left. Then (recall that the integral with respect to a Dirac measure is simply the value of the function at the point)

  \[ \begin{align} \int _{\mathbb {R}} \sqrt{\left|x + 2\right|} \; d\mu _{s} & = 0.1 \cdot \sqrt{\left|-2 + 2\right|} + 0.1 \cdot \sqrt{\left|-1 + 2\right|} + 0.05 \cdot \sqrt{\left|\sqrt[3]{2} + 2\right|} +\\ & \qquad + 0.05 \cdot \sqrt{\left|5 + 2\right|} + 0.1 \cdot \underbrace{\int \sqrt{\left|x+2\right|} \; d\mu _{C-2}}_{=\int _{[0,1]}\sqrt{x} \; d\mu _{c} \approx 0.64} \approx \\ & \approx 0.322 + 0.064 = 0.386 \end{align} \]

  We omit the detailed computation of the integrals with respect to the Cantor measure, since that is not the point of the example; what matters is understanding the general principle.

In summary, \(\mathbb {E}\left[\sqrt{\left|X + 2\right|}\right] \approx 1.107 + 0.386 = 1.493\)

ProblemProblem 13

Consider the function \[ F(x)= \begin{cases} \frac{1}{2}x-1, & x<0,\\ x^2, & 0 \le x < 1,\\ 3, & 1 \le x < 2,\\ 6 - \dfrac {4}{x}, & x \ge 2. \end{cases} \]

It is right-continuous. Hence one can construct from it a Lebesgue-Stieltjes measure \(\mu_F\).

1. Decompose \(\mu_F\) into an absolutely continuous part \(\mu_a\) and a singular part \(\mu_s\) with respect to Lebesgue measure;

2. For the absolutely continuous part \(\mu_a\) find the density (Radon-Nikodym derivative) \(\frac{d\mu_a}{d\lambda }\);

3. Is the singular part \(\mu_s\) (relative to Lebesgue measure) absolutely continuous with respect to counting measure \(\#_{\mathbb {N}}\) on the natural numbers? What about with respect to \(\#_{\mathbb {Z}}\) on the integers? If so, find the Radon-Nikodym derivative \(\frac{d\mu_s}{d\# }\).

4. Find the integral

\[ \int _{[-2, +\infty ]} \sqrt{\left|x\right|} \; \; d\mu _F \]

4 (*) Density of a set at a point

Definition 3 Let \(E \subseteq \mathbb {R}\). The density of the set \(E\) at a point \(b \in \mathbb {R}\) is the limit \[ \lim _{t \downarrow 0} \frac{\lambda \left(E \cap (b-t, \, b+t)\right)}{2t}, \] if it exists (otherwise the density of \(E\) at \(b\) is undefined).

For example, the density of the segment \([0,1]\) equals \(1\) at points \(b \in (0,1)\), equals \(\frac{1}{2}\) at points \(b \in \left\{ 0, 1\right\}\), and equals \(0\) at all other points.

Theorem 5 (Lebesgue density theorem) Let \(E \subseteq \mathbb {R}\) be Lebesgue measurable. Then the density of \(E\) equals \(1\) at almost every point of \(E\) and equals \(0\) at almost every point of \(\mathbb {R} \setminus E\).

ProblemProblem 14

Give an example of a Borel set \(E \subseteq \mathbb {R}\) whose density at zero

1. is undefined;

2. equals \(\frac{1}{3}\);

3. equals a prescribed number \(s \in [0,1]\).

ProblemProblem 15

Let \(E \subseteq \mathbb {R}\) be Lebesgue measurable, with the density of \(E\) equal to \(1\) at every point of \(E\) and equal to \(0\) at every point of \(\mathbb {R} \setminus E\) — with no exceptions of measure zero. Prove that \(E = \varnothing\) or \(E = \mathbb {R}\).

HintHint

Assume that both sets \(E\) and \(\mathbb {R} \setminus E\) are nonempty. Take \(a \in E\), \(b \not\in E\) and consider the “boundary” point between them; show that the density there can equal neither \(0\) nor \(1\).

Let us end with a striking contrast to the density theorem.

Theorem 6 (A “bad” Borel set) There exists a Borel set \(E \subseteq \mathbb {R}\) such that \[ 0 < \lambda (E \cap I) < \lambda (I) \] for every nonempty bounded interval \(I\).

Before reading the proof, consider why this does not contradict the Lebesgue density theorem (Theorem 5): at small scales around almost every one of its points, \(E\) occupies almost all the space — and yet no interval is filled entirely by the set \(E\), and none is left entirely free of it. Compare this also with the example above about “half of every interval”: the requirement \(\lambda (E \cap I) = \frac{\lambda (I)}{2}\) is impossible to satisfy, whereas \(0 < \lambda (E \cap I) < \lambda (I)\) is entirely achievable.

Mn-remarkRemark

The set \(E\) from theorem Theorem 6 also gives the promised “incurably discontinuous” function: \(\; \mathbb {1}_{E}\) is discontinuous at every point, and no modification of \(\; \mathbb {1}_{E}\) on a set of measure zero will make it continuous at even a single point (check this: in any interval, both the set \(E\) and its complement have positive measure, so a correction on a set of measure zero does not change this). Compare with \(\; \mathbb {1}_{\mathbb {Q}}\): it too is discontinuous everywhere, but after modification on a set of measure zero — on \(\mathbb {Q}\) itself — it becomes identically zero. At the same time the function \[ g(b) = \int _{0}^{b}\; \mathbb {1}_{E} \] is, by Lebesgue’s theorem on differentiation of the integral (Theorem 3), differentiable almost everywhere: differentiability of the “antiderivative” requires no continuity whatsoever of the integrand.

5 A summary map: classes of measures and absolute continuity

Let us gather together all the classes of (nonnegative) measures on \(\mathbb {R}\) encountered in this chapter, and how absolute continuity with respect to \(\lambda\) intersects them.

Classes of nonnegative measures on \(\mathbb{R}\) — nested rings. The vertical band crossing all the rings is the absolutely continuous (with respect to Lebesgue measure \(\lambda\)) measures: the intersection of the band with each ring is the class of AC measures at that level, and one can see that each successive such class is nested inside the previous one. Click “statements ▸” in any compartment of the band to expand the equivalent descriptions.

Absolutely continuous (w.r.t. Lebesgue measure \(\lambda\)) measures \(\mu\)

All (nonnegative) measures \(\mu\) on \((\mathbb{R},\mathscr{B}(\mathbb{R}))\)

example outside the band: counting measure on \(\mathbb{R}\) — not \(\sigma\)-finite and not AC

AC \(\sigma\)-finite measures

\(\sigma\)-finite measures

example outside the band: counting measure on \(\mathbb{Q}\) — \(\sigma\)-finite, but not locally finite and not AC

AC locally finite measures

Locally finite measures = Lebesgue–Stieltjes measures \(\mu_F\) (\(F\), respectively, a nondecreasing cadlag function on \(\mathbb{R}\))

AC finite measures

Finite measures = L-S measures \(\mu_F\) for which \(F\) is bounded

AC probability measures

Probability measures = finite measures with \(\mu(\mathbb{R})=1\) = L-S measures \(\mu_F\) for which \(\sup\operatorname{range}F - \inf\operatorname{range}F = 1\)

examples outside the band: \(\delta_0\) (an atom at zero), the Cantor measure (continuous, but singular)

AC band: μ ≪ λ

Mn-remarkRemark

5.1 a dictionary: \(\mu \leftrightarrow F \leftrightarrow f\)

The diagram above actually encodes not only classes of measures \(\mu\), but also the fact that each of them is bijectively arranged also as a class of functions \(F\) and/or as a class of densities \(f\).

• \(\mu \leftrightarrow F\). The correspondence from lemma Lemma 3 is a bijection between locally finite measures \(\mu\) and equivalence classes of nondecreasing right-continuous functions \(F: \mathbb {R} \to \mathbb {R}\) under the relation \(F \sim G \iff F - G \equiv \mathrm{const}\) (indeed: \(\mu_F = \mu_G \iff F(b)-F(a) = G(b)-G(a) \; \forall a<b \iff F-G \equiv \mathrm{const}\)). Finite measures correspond exactly to bounded such \(F\), again up to a constant. Outside the class of locally finite measures, such an (everywhere finite) function \(F\) simply does not exist — which is why the two outer rings of the diagram (arbitrary measures and merely \(\sigma\)-finite measures) carry no label with \(F\).

• \(\mu \leftrightarrow f\). The correspondence \(\mu \leftrightarrow f\) is a bijection between measures \(\mu\) that are absolutely continuous (with respect to \(\lambda\)) and equivalence classes of measurable functions \(f: \mathbb {R} \to [0,+\infty ]\) under the relation \(f \sim g \iff f=g\) \(\lambda\)-a.e. Within the band of absolute continuity, \(\sigma\)-finiteness of \(\mu\) corresponds to finiteness of \(f\) \(\lambda\)-a.e., local finiteness to \(f \in L^1_{loc}(\mathbb {R})\), and finiteness to \(f \in L^1(\mathbb {R})\) (everywhere, up to agreement \(\lambda\)-a.e.).

• \(F \leftrightarrow f\). Where both correspondences are defined — i.e. for absolutely continuous locally finite measures — they are consistent by theorem Theorem 4: \(F' = f\) \(\lambda\)-a.e.

For example, the innermost lens of the diagram (finite \(\cap\) absolutely continuous measures) is simultaneously three descriptions of the same set of objects: finite absolutely continuous measures \(\mu\); bounded locally absolutely continuous functions \(F\) (up to an additive constant); functions \(f \in L^1(\mathbb {R})\), \(f \geq 0\) (up to agreement \(\lambda\)-a.e.).