Measures and Integration

Integration theory rests on a single move. One defines the integral of a nonnegative function as the supremum of the integrals of the simple functions beneath it, then proves that this construction commutes with monotone limits. The standard interchange theorems used in probability descend from that one theorem.

#Measure spaces

Definition1

A $\sigma$ -algebra on a set $\Omega$ is a collection $\mathcal F$ of subsets containing $\Omega$ and closed under complementation and countable unions. A measure is a function $\mu:\mathcal F\to[0,\infty]$ with $\mu(\varnothing)=0$ that is countably additive, so that for pairwise disjoint $A_1,A_2,\dots\in\mathcal F$ ,

\mu\!\Big(\bigcup_{n} A_n\Big)=\sum_{n}\mu(A_n). \tag{1}

The triple $(\Omega,\mathcal F,\mu)$ is a measure space, and $\mu$ is a probability measure when $\mu(\Omega)=1$ .

A function $f:\Omega\to\R$ is measurable when $f^{-1}((a,\infty))\in\mathcal F$ for every $a\in\R$ , equivalently when preimages of Borel sets lie in $\mathcal F$ . The same definition applies to $[0,\infty]$ -valued functions, with $f$ measurable when $f^{-1}((a,\infty])\in\mathcal F$ for every $a\in\R$ ; the supremum integral below and the monotone limits we build are taken in this extended-real class. Measurable functions are closed under pointwise limits because $\{\sup_n f_n>a\}=\bigcup_n\{f_n>a\}$ and $\{\inf_n f_n<a\}=\bigcup_n\{f_n<a\}$ exhibit $\sup$ , $\inf$ , and hence $\limsup$ , $\liminf$ , $\lim$ as countable set operations. Closure under sums follows from $\{f+g>a\}=\bigcup_{r\in\mathbb Q}(\{f>r\}\cap\{g>a-r\})$ , a countable union of measurable sets, and closure under products from $fg=\tfrac12((f+g)^2-f^2-g^2)$ with $x\mapsto x^2$ continuous. Existence of nontrivial measures rests on the extension theorem below.

Theorem2

(Caratheodory extension.) A countably additive premeasure on an algebra of subsets of $\Omega$ extends to a measure on the generated $\sigma$ -algebra, and the extension is unique when the premeasure is $\sigma$ -finite. Lebesgue measure on $\R$ is the extension of the interval-length premeasure defined on the algebra of finite unions of intervals [1], [2].

#The Lebesgue integral

A simple function is a finite combination $s=\sum_{i=1}^m a_i\,\ind_{A_i}$ with $a_i\ge 0$ and $A_i\in\mathcal F$ , and its integral is $\int s\,d\mu=\sum_{i=1}^m a_i\,\mu(A_i)$ , a value independent of the representation. For measurable $f\ge 0$ define

\int f\,d\mu=\sup\Big\{\textstyle\int s\,d\mu : s\text{ simple},\ 0\le s\le f\Big\}, \tag{2}

and for general measurable $f$ , write $f=f^+-f^-$ with $f^\pm=\max(\pm f,0)$ and set $\int f\,d\mu=\int f^+\,d\mu-\int f^-\,d\mu$ whenever the difference is not $\infty-\infty$ . The function $f$ is integrable, $f\in L^1(\mu)$ , when $\int\abs{f}\,d\mu<\infty$ .

#The convergence theorems

Theorem3

(Monotone convergence.) If $f_n\ge 0$ are measurable with $f_n\uparrow f$ pointwise, then $\int f_n\,d\mu\uparrow\int f\,d\mu$ .

Proof

The sequence $\int f_n\,d\mu$ is nondecreasing and bounded above by $\int f\,d\mu$ from Equation (2), so it converges to some $L\le\int f\,d\mu$ . Fix a simple $s$ with $0\le s\le f$ and a constant $c\in(0,1)$ , and set $E_n=\{f_n\ge cs\}$ . Each $E_n\in\mathcal F$ , the $E_n$ increase, and $\bigcup_n E_n=\Omega$ . Where $s>0$ one has $f\ge s>cs$ and $f_n\uparrow f$ , so $f_n\ge cs$ eventually, while where $s=0$ one has $cs=0\le f_n$ for every $n$ . Then

\int f_n\,d\mu\ge\int_{E_n} f_n\,d\mu\ge c\int_{E_n} s\,d\mu, \tag{3}

and continuity from below of the measure $A\mapsto\int_A s\,d\mu$ , itself a consequence of Equation (1), gives $\int_{E_n} s\,d\mu\uparrow\int s\,d\mu$ . Letting $n$ grow yields $L\ge c\int s\,d\mu$ ; letting $c$ approach $1$ and taking the supremum over $s$ gives $L\ge\int f\,d\mu$ . Both inequalities hold, so $L=\int f\,d\mu$ .

Corollary4

(Fatou's lemma.) For measurable $f_n\ge 0$ , $\int\liminf_n f_n\,d\mu\le\liminf_n\int f_n\,d\mu$ .

Proof

Put $g_n=\inf_{k\ge n} f_k$ . Then $g_n$ is measurable, $0\le g_n\uparrow\liminf_n f_n$ , and $g_n\le f_k$ for every $k\ge n$ , so $\int g_n\,d\mu\le\inf_{k\ge n}\int f_k\,d\mu$ . Applying Theorem 3 to $g_n$ on the left and passing to the limit on the right delivers the claim.

Theorem5

(Dominated convergence.) Suppose measurable $f_n$ satisfy $f_n\to f$ pointwise and $\abs{f_n}\le g$ for a fixed $g\in L^1(\mu)$ . Then $f\in L^1(\mu)$ and $\int\abs{f_n-f}\,d\mu$ vanishes; in particular $\int f_n\,d\mu$ converges to $\int f\,d\mu$ .

Proof

The bound $\abs{f}\le g$ passes to the limit, so $f\in L^1(\mu)$ . The functions $2g-\abs{f_n-f}$ are nonnegative, so Corollary 4 gives

\int 2g\,d\mu\le\liminf_n\int\big(2g-\abs{f_n-f}\big)\,d\mu=\int 2g\,d\mu-\limsup_n\int\abs{f_n-f}\,d\mu. \tag{4}

Since $\int 2g\,d\mu$ is finite it cancels, leaving $\limsup_n\int\abs{f_n-f}\,d\mu\le 0$ . The integrand is nonnegative, so the limit superior is exactly $0$ , and $\abs{\int f_n\,d\mu-\int f\,d\mu}\le\int\abs{f_n-f}\,d\mu$ closes the argument.

#The $L^p$ inequalities

For $1\le p<\infty$ let $\norm{f}_p=(\int\abs{f}^p\,d\mu)^{1/p}$ and let $L^p(\mu)$ collect the measurable $f$ with $\norm{f}_p<\infty$ , identified up to equality almost everywhere.

Proposition6

(Holder and Minkowski.) For conjugate exponents $1<p,q<\infty$ with $1/p+1/q=1$ , $\int\abs{fg}\,d\mu\le\norm{f}_p\norm{g}_q$ , and for $1\le p<\infty$ , $\norm{f+g}_p\le\norm{f}_p+\norm{g}_p$ .

Proof

For Holder, if $\norm f_p=\infty$ or $\norm g_q=\infty$ the bound is trivial, and if $\norm f_p=0$ then $f=0$ almost everywhere so $\int\abs{fg}=0$ , likewise for $\norm g_q=0$ . In the remaining case $0<\norm f_p,\norm g_q<\infty$ , Young's inequality $ab\le a^p/p+b^q/q$ follows from concavity of the logarithm. Apply it to $a=\abs f/\norm f_p$ and $b=\abs g/\norm g_q$ and integrate, which gives $\int\abs{fg}/(\norm f_p\norm g_q)\le 1/p+1/q=1$ . For Minkowski the case $p=1$ is immediate from $\abs{f+g}\le\abs f+\abs g$ integrated, so assume $1<p<\infty$ . If $\norm f_p=\infty$ or $\norm g_p=\infty$ the bound is trivial, and if $\norm{f+g}_p=0$ it is trivial. Otherwise convexity of $t\mapsto t^p$ gives $\abs{f+g}^p\le 2^{p-1}(\abs f^p+\abs g^p)$ , so $\norm{f+g}_p<\infty$ is finite and nonzero. Write $\abs{f+g}^p\le\abs{f+g}^{p-1}(\abs f+\abs g)$ , integrate, and apply Holder to each term with exponent $q=p/(p-1)$ , using $\int(\abs{f+g}^{p-1})^q=\int\abs{f+g}^p<\infty$ , to obtain $\norm{f+g}_p^p\le\norm{f+g}_p^{p-1}(\norm f_p+\norm g_p)$ ; dividing by the finite nonzero $\norm{f+g}_p^{p-1}$ yields the triangle inequality.

Theorem7

(Riesz-Fischer.) Each $L^p(\mu)$ is complete. Every Cauchy sequence converges in $\norm{\cdot}_p$ to an element of $L^p(\mu)$ [1].

#A limit that is not an integral

Domination in Theorem 5 cannot be dropped. On $(0,1)$ with Lebesgue measure the bumps $f_n=n\,\ind_{(0,1/n)}$ converge pointwise to $0$ , yet each has integral $1$ ; no integrable function dominates the family, and the limit of the integrals is not the integral of the limit.

import numpy as np


def escaping_mass_integral(n: int, grid: int = 1_000_000) -> float:
    """Midpoint estimate of the integral of n * indicator(0, 1/n) on (0, 1).

    Args:
        n: Height and inverse width of the bump.
        grid: Number of equal subintervals discretizing the unit interval.

    Returns:
        The approximate integral, equal to one for every n.
    """
    midpoints = (np.arange(grid) + 0.5) / grid
    heights = np.where(midpoints < 1.0 / n, float(n), 0.0)
    return float(heights.mean())


integrals = [escaping_mass_integral(n) for n in (1, 10, 100, 1_000)]
pointwise_limit = 0.0

The monotone convergence theorem is the one result that survives without a dominating function, and it is the engine from which the other two were derived.

[1]

G. B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed. Wiley, 1999.

[2]

P. Billingsley, Probability and Measure, 3rd ed. Wiley, 1995.

Explore connections

see in the atlas

referenced by (11)

cite

@misc{measures-and-integration,
  author = {Zac Kienzle},
  title  = {Measures and Integration},
  year   = {2026},
  month  = {06},
  url    = {https://zackienzle.com/blog/measures-and-integration}
}

#Measure spaces

Definition1

\mu\!\Big(\bigcup_{n} A_n\Big)=\sum_{n}\mu(A_n). \tag{1}

The triple $(\Omega,\mathcal F,\mu)$ is a measure space, and $\mu$ is a probability measure when $\mu(\Omega)=1$ .

Theorem2

#The Lebesgue integral

\int f\,d\mu=\sup\Big\{\textstyle\int s\,d\mu : s\text{ simple},\ 0\le s\le f\Big\}, \tag{2}

#The convergence theorems

Theorem3

(Monotone convergence.) If $f_n\ge 0$ are measurable with $f_n\uparrow f$ pointwise, then $\int f_n\,d\mu\uparrow\int f\,d\mu$ .

Proof

\int f_n\,d\mu\ge\int_{E_n} f_n\,d\mu\ge c\int_{E_n} s\,d\mu, \tag{3}

Corollary4

(Fatou's lemma.) For measurable $f_n\ge 0$ , $\int\liminf_n f_n\,d\mu\le\liminf_n\int f_n\,d\mu$ .

Proof

Theorem5

Proof

The bound $\abs{f}\le g$ passes to the limit, so $f\in L^1(\mu)$ . The functions $2g-\abs{f_n-f}$ are nonnegative, so Corollary 4 gives

\int 2g\,d\mu\le\liminf_n\int\big(2g-\abs{f_n-f}\big)\,d\mu=\int 2g\,d\mu-\limsup_n\int\abs{f_n-f}\,d\mu. \tag{4}

#The $L^p$ inequalities

For $1\le p<\infty$ let $\norm{f}_p=(\int\abs{f}^p\,d\mu)^{1/p}$ and let $L^p(\mu)$ collect the measurable $f$ with $\norm{f}_p<\infty$ , identified up to equality almost everywhere.

Proposition6

(Holder and Minkowski.) For conjugate exponents $1<p,q<\infty$ with $1/p+1/q=1$ , $\int\abs{fg}\,d\mu\le\norm{f}_p\norm{g}_q$ , and for $1\le p<\infty$ , $\norm{f+g}_p\le\norm{f}_p+\norm{g}_p$ .

Proof

Theorem7

(Riesz-Fischer.) Each $L^p(\mu)$ is complete. Every Cauchy sequence converges in $\norm{\cdot}_p$ to an element of $L^p(\mu)$ [1].

#A limit that is not an integral

import numpy as np


def escaping_mass_integral(n: int, grid: int = 1_000_000) -> float:
    """Midpoint estimate of the integral of n * indicator(0, 1/n) on (0, 1).

    Args:
        n: Height and inverse width of the bump.
        grid: Number of equal subintervals discretizing the unit interval.

    Returns:
        The approximate integral, equal to one for every n.
    """
    midpoints = (np.arange(grid) + 0.5) / grid
    heights = np.where(midpoints < 1.0 / n, float(n), 0.0)
    return float(heights.mean())


integrals = [escaping_mass_integral(n) for n in (1, 10, 100, 1_000)]
pointwise_limit = 0.0

The monotone convergence theorem is the one result that survives without a dominating function, and it is the engine from which the other two were derived.

[1]

G. B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed. Wiley, 1999.

[2]

P. Billingsley, Probability and Measure, 3rd ed. Wiley, 1995.

Explore connections

see in the atlas

referenced by (11)

cite

@misc{measures-and-integration,
  author = {Zac Kienzle},
  title  = {Measures and Integration},
  year   = {2026},
  month  = {06},
  url    = {https://zackienzle.com/blog/measures-and-integration}
}

#Measure spaces

#The Lebesgue integral

#The convergence theorems

#The LpL^pLp inequalities

#A limit that is not an integral

#Measure spaces

#The Lebesgue integral

#The convergence theorems

#The LpL^pLp inequalities

#A limit that is not an integral

#The $L^p$ inequalities

#The $L^p$ inequalities