Measures and Measurable Functions

Size is encoded by a measure space $(X,\Sigma,\mu)$: the sigma-algebra $\Sigma$ selects the subsets whose size is defined, and the countably additive map $\mu$ assigns that size. A statement holds $\mu$-almost everywhere when its failures lie inside a $\mu$-null set. A measurable observable is a map whose output questions pull back to sets in $\Sigma$; it transports $\mu$ to a measure on its value space.

This page builds those structures and tests them on a normalized, finite-dimensional Euclidean Gaussian. Familiar finite-dimensional integrals appear only as checks; the general construction of the Lebesgue integral, convergence theorems, product measures, and continuum field measures is developed on the linked pages.

Measurable spaces · Measures · Null sets · Measurable maps · Pushforwards · Gaussian example · Exercises

The data of a measure space

The central objects have different mathematical types.

Data	Type	Role
$X$	set	possible points, outcomes, or configurations
$\Sigma\subseteq\mathcal P(X)$	sigma-algebra	subsets that may be measured
$\mu:\Sigma\to[0,\infty]$	measure	countably additive size
$f:(X,\Sigma)\to(Y,\mathcal T)$	measurable map	observable or change of description
$f_{\#}\mu$	measure on $(Y,\mathcal T)$	distribution of observable values

Neither $\Sigma$ nor $\mu$ is implicit. The same set $X$ can carry different sigma-algebras, and the same measurable space can carry different measures. Consequently, the phrases “measurable,” “null,” and “almost everywhere” must always be interpreted relative to named data.

Sigma-algebras select the measurable sets

Let $X$ be a set. A sigma-algebra $\Sigma$ on $X$ is a collection of subsets satisfying

$$\varnothing\in\Sigma, \qquad A\in\Sigma\Longrightarrow X\setminus A\in\Sigma, \qquad A_1,A_2,\ldots\in\Sigma \Longrightarrow \bigcup_{n=1}^{\infty}A_n\in\Sigma.$$

The pair $(X,\Sigma)$ is a measurable space, and the members of $\Sigma$ are its measurable sets. De Morgan’s laws give closure under countable intersections; padding with empty sets gives finite unions; complements then give differences. Countable closure is what later keeps level sets measurable under limits.

Two extremes are always available:

$$\{\varnothing,X\} \qquad\text{and}\qquad \mathcal P(X),$$

the trivial and discrete sigma-algebras. Between them, a sigma-algebra records which distinctions the theory can make.

A finite example and a near miss

Let $X=\{a,b,c,d\}$ and suppose the only primitive distinction is between the two blocks $\{a,b\}$ and $\{c,d\}$. Then

$$\Sigma = \{\varnothing,\{a,b\},\{c,d\},X\}$$

is a sigma-algebra. It can ask whether a point lies in either block, but it cannot ask whether the point is exactly $a$.

By contrast, on $X=\{a,b,c\}$ the collection

$$\mathcal C=\{\varnothing,\{a\},\{b\},X\}$$

is not a sigma-algebra: the complement $X\setminus\{a\}=\{b,c\}$ is absent. A proposed collection must satisfy every axiom, not merely contain $\varnothing$ and $X$.

Generated and Borel sigma-algebras

For any family $\mathcal G\subseteq\mathcal P(X)$, the generated sigma-algebra

$$\sigma(\mathcal G) = \bigcap\{\Sigma':\Sigma'\text{ is a sigma-algebra and } \mathcal G\subseteq\Sigma'\}$$

is the smallest sigma-algebra containing $\mathcal G$. Thus the finite example above is $\sigma(\{\{a,b\}\})$.

If $X$ is a topological space, its Borel sigma-algebra is

$$\mathcal B(X)=\sigma(\{U\subseteq X:U\text{ is open}\}).$$

On $\mathbb R$, the same sigma-algebra is generated by open intervals, closed intervals, or rays such as $(a,\infty)$. On $\mathbb R^d$, open sets or half-open rectangles generate $\mathcal B(\mathbb R^d)$. This flexibility is important: to prove that a map is measurable, it is enough to check a family that generates the target sigma-algebra; see Melrose 2004, §§ 2–3, PDF.

Measures and countable additivity

A measure on $(X,\Sigma)$ is a map

$$\mu:\Sigma\longrightarrow[0,\infty]$$

such that $\mu(\varnothing)=0$ and, for every pairwise disjoint sequence $(A_n)$ in $\Sigma$,

$$\mu\!\left(\bigcup_{n=1}^{\infty}A_n\right) = \sum_{n=1}^{\infty}\mu(A_n).$$

The triple $(X,\Sigma,\mu)$ is a measure space. Allowing the value $\infty$ is essential: Lebesgue measure of all of $\mathbb R^d$ and counting measure of an infinite set are both infinite.

A measure is finite if $\mu(X)<\infty$, a probability measure if $\mu(X)=1$, and sigma-finite if

$$X=\bigcup_{n=1}^{\infty}X_n \qquad\text{with}\qquad X_n\in\Sigma, \quad \mu(X_n)<\infty.$$

Sigma-finite does not mean finite. Lebesgue measure on $\mathbb R^d$ is sigma-finite because the space is a countable union of bounded cubes, but its total mass is infinite.

Standard examples include:

• Counting measure: $\#A$ is the number of elements of $A$, possibly infinite.
• Dirac measure: $\delta_x(A)=1$ if $x\in A$ and $0$ otherwise.
• Weighted discrete measure: on a countable set, $\mu(A)=\sum_{x\in A}w_x$ for weights $w_x\geq0$.
• Lebesgue measure: $\lambda^d$ assigns ordinary volume to suitable subsets of $\mathbb R^d$.

Consequences and their hypotheses

Countable additivity yields the following working rules.

Statement	Exact hypothesis
$A\subseteq B\Rightarrow\mu(A)\leq\mu(B)$	$A,B\in\Sigma$
$\mu(\bigcup_n A_n)\leq\sum_n\mu(A_n)$	every $A_n\in\Sigma$
$A_n\uparrow A\Rightarrow\mu(A_n)\to\mu(A)$	every $A_n\in\Sigma$
$A_n\downarrow A\Rightarrow\mu(A_n)\to\mu(A)$	every $A_n\in\Sigma$ and $\mu(A_1)<\infty$

Here $A_n\uparrow A$ means $A_n\subseteq A_{n+1}$ and $A=\bigcup_nA_n$; similarly, $A_n\downarrow A$ means decreasing sets with $A=\bigcap_nA_n$.

For continuity from below, disjointize the growth:

$$B_1=A_1, \qquad B_n=A_n\setminus A_{n-1}\quad(n\geq2).$$

Then $A=\bigsqcup_nB_n$ and $A_N=\bigsqcup_{n\leq N}B_n$, so countable additivity gives

$$\mu(A) = \sum_{n=1}^{\infty}\mu(B_n) = \lim_{N\to\infty}\sum_{n=1}^{N}\mu(B_n) = \lim_{N\to\infty}\mu(A_N).$$

For continuity from above, apply this result to $A_1\setminus A_n\uparrow A_1\setminus A$. The assumption $\mu(A_1)<\infty$ permits subtraction from $\mu(A_1)$ without an $\infty-\infty$ ambiguity. It is enough that some $A_n$ have finite measure, because the sequence may then be reindexed.

The finite-measure condition cannot simply be dropped. With Lebesgue measure on $\mathbb R$,

$$A_n=[n,\infty) \downarrow \varnothing,$$

but $\lambda(A_n)=\infty$ for every $n$, whereas $\lambda(\varnothing)=0$.

Borel sets, Lebesgue measure, and completion

Lebesgue measure is not defined on every subset of $\mathbb R^d$. One route to its domain begins with the outer measure

$$\lambda^{d,*}(E) = \inf_{E\subseteq\bigcup_{j=1}^{\infty}Q_j} \sum_{j=1}^{\infty}|Q_j|,$$

where the infimum runs over countable box covers and $|Q_j|$ is Euclidean box volume. A set $E$ passes the Carathéodory test when

$$\lambda^{d,*}(A) = \lambda^{d,*}(A\cap E) + \lambda^{d,*}(A\setminus E) \qquad\text{for every }A\subseteq\mathbb R^d.$$

The sets that pass form the Lebesgue sigma-algebra, and restricting outer measure to it gives the complete Lebesgue measure $\lambda^d$. In particular,

$$\lambda^d\!\left(\prod_{j=1}^{d}(a_j,b_j]\right) = \prod_{j=1}^{d}(b_j-a_j).$$

There are three different domains to keep separate:

1. $\mathcal B(\mathbb R^d)$ contains the sets generated by the open sets.
2. The Lebesgue sigma-algebra completes the Borel measure by adding every subset of every Borel null set, and sets obtained from these by measurable operations.
3. $\mathcal P(\mathbb R^d)$ contains all subsets and is strictly larger.

The restriction is forced by the desired properties of length. To see the obstruction, use a choice of one representative from every equivalence class of $[0,1]$ under $x\sim y$ when $x-y\in\mathbb Q$, and call the resulting set $V$. The translates $V+q$ for $q\in\mathbb Q\cap[-1,1]$ are disjoint, their union contains $[0,1]$, and that union lies in $[-1,2]$. If a translation- invariant, countably additive extension of interval length measured every subset, then $\lambda(V)=0$ would make the union have measure zero, while $\lambda(V)>0$ would make it have infinite measure. Both contradict the two interval bounds. Thus not every subset can be assigned such a length.

Null sets and almost-everywhere statements

A null set is a measurable set $N\in\Sigma$ with $\mu(N)=0$. A property $P(x)$ holds $\mu$-almost everywhere, abbreviated $\mu$-a.e., if there is a measurable null set $N$ such that $P(x)$ holds for every $x\in X\setminus N$. The measure must be named: the same exceptional set can be null for one measure and have positive mass for another.

For example, $\mathbb Q\cap[0,1]$ is dense in $[0,1]$ but has Lebesgue measure zero. It has Dirac mass one under $\delta_0$. Null therefore does not mean empty, impossible, finite, or topologically small.

A measure space is complete if every subset of every measurable null set is itself measurable and null. This is unrelated to metric completeness: one concerns subsets of measure zero, while the other concerns limits of Cauchy sequences.

The distinction is visible on a three-point space. Let

$$X=\{a,b,c\}, \qquad \Sigma=\{\varnothing,\{a,b\},\{c\},X\},$$

and define

$$\mu(\varnothing)=\mu(\{a,b\})=0, \qquad \mu(\{c\})=\mu(X)=1.$$

Then $\{a,b\}$ is null, but its subset $\{a\}$ is not measurable. The completion enlarges $\Sigma$ by all subsets of null sets and their measurable unions; here it produces $\mathcal P(X)$. This is why changing a measurable function on an arbitrary subset of a null set can destroy measurability in an incomplete space.

For measurable real-valued functions $f$ and $g$, the statement $f=g$ $\mu$-a.e. means

$$\mu(\{x\in X:f(x)\neq g(x)\})=0.$$

Likewise, $f_n\to f$ $\mu$-a.e. means that pointwise convergence holds away from one measurable null set. This is weaker than pointwise convergence everywhere and is different for different measures.

Measurable maps and observables

Let $(X,\Sigma)$ and $(Y,\mathcal T)$ be measurable spaces. A map

$$f:(X,\Sigma)\longrightarrow(Y,\mathcal T)$$

is measurable when

$$f^{-1}(B)\in\Sigma \qquad\text{for every }B\in\mathcal T.$$

Inverse images are the correct operation because they preserve exactly the set operations used by sigma-algebras:

$$\begin{aligned} f^{-1}(Y\setminus B)&=X\setminus f^{-1}(B),\\ f^{-1}\!\left(\bigcup_nB_n\right)&=\bigcup_nf^{-1}(B_n). \end{aligned}$$

If $\mathcal T=\sigma(\mathcal G)$, it is enough to verify $f^{-1}(G)\in\Sigma$ for each $G\in\mathcal G$. Indeed, the target sets with measurable inverse image themselves form a sigma-algebra containing $\mathcal G$.

For $f:X\to[-\infty,\infty]$, measurability is equivalent to

$$\{x:f(x)>a\}\in\Sigma \qquad\text{for every }a\in\mathbb R,$$

because these rays generate the Borel sigma-algebra of the extended real line. Several useful consequences follow.

• Every continuous map between topological spaces is measurable for their Borel sigma-algebras.
• The indicator $\mathbf1_A:X\to\{0,1\}$, with $\{0,1\}$ given its discrete sigma-algebra, is measurable exactly when $A\in\Sigma$.
• Compositions of measurable maps are measurable, since $(g\circ f)^{-1}(C)=f^{-1}(g^{-1}(C))$.
• Sums and products of real- or complex-valued measurable functions are measurable wherever the corresponding extended-real operations are defined.

The finite incomplete space above supplies a nonexample. Give $\{0,1\}$ its discrete sigma-algebra and set $f(a)=1$, $f(b)=f(c)=0$. Then $f^{-1}(\{1\})=\{a\}\notin\Sigma$, so $f$ is not measurable. The formula for the map is not enough; its inverse images must belong to the chosen domain.

Pointwise limits and simple functions

Measurability is stable under countable limiting operations. Suppose $f_n:X\to\mathbb R$ are measurable and $f_n(x)\to f(x)$ for every $x$. For any $a\in\mathbb R$,

$$\{f>a\} = \bigcup_{\substack{r\in\mathbb Q\\r>a}} \bigcup_{N=1}^{\infty} \bigcap_{n\geq N}\{f_n>r\}.$$

The right-hand side uses only countable unions and intersections of measurable sets. Hence $f$ is measurable. The same level-set reasoning shows that countable suprema, infima, limsups, and liminfs of measurable extended-real functions are measurable. None of these statements says that an integral may be interchanged with a limit; that requires additional hypotheses on the next page. For the pointwise-limit closure argument, see Lin 2021, Lecture 9, PDF.

A measurable simple function has finite range and can be written

$$s=\sum_{j=1}^{m}a_j\mathbf1_{A_j}, \qquad A_j\in\Sigma.$$

Every nonnegative measurable $f:X\to[0,\infty]$ is an increasing pointwise limit of nonnegative measurable simple functions. One explicit construction is

$$s_n(x)= \begin{cases} 2^{-n}\lfloor2^nf(x)\rfloor,&f(x)<2^n,\\ 2^n,&f(x)\geq2^n. \end{cases}$$

Each $s_n$ has finite range, $0\leq s_n\leq s_{n+1}\leq f$, and $s_n(x)\to f(x)$. If $f(x)<\infty$, eventually the truncation is inactive and $0\leq f(x)-s_n(x)<2^{-n}$; if $f(x)=\infty$, then $s_n(x)=2^n$. The Lebesgue integration page uses these approximants to define the integral and prove convergence theorems.

Pushforward measures and distributions

Let $f:(X,\Sigma)\to(Y,\mathcal T)$ be measurable and let $\mu$ be a measure on $(X,\Sigma)$. The pushforward of $\mu$ by $f$ is

$$f_{\#}\mu(B) = \mu(f^{-1}(B)), \qquad B\in\mathcal T.$$

The notation $f_*\mu$ is also common. Measurability ensures that the right- hand side is defined. If $(B_n)$ are pairwise disjoint target sets, then their inverse images are pairwise disjoint and

$$\begin{aligned} f_{\#}\mu\!\left(\bigcup_nB_n\right) &=\mu\!\left(f^{-1}\!\left(\bigcup_nB_n\right)\right)\\ &=\mu\!\left(\bigcup_nf^{-1}(B_n)\right)\\ &=\sum_n\mu(f^{-1}(B_n)) =\sum_nf_{\#}\mu(B_n). \end{aligned}$$

Thus $f_{\#}\mu$ is a measure, and its total mass is preserved:

$$(f_{\#}\mu)(Y)=\mu(X).$$

When $\mu$ is a probability measure, $f_{\#}\mu$ is the law or distribution of the observable $f$. A constant map $f(x)=y_0$ sends a probability measure to $\delta_{y_0}$.

A measure is not merely a density

A nonnegative density must be measurable relative to a named reference measure. The notation

$$\mathrm d\nu(x)=\rho(x)\,\mathrm d\lambda(x)$$

means that $\nu(A)=\int_A\rho\,\mathrm d\lambda$ for measurable $A$; the integration theory is developed next. Changing the reference measure or coordinates changes the density even when the underlying measure is the same. Moreover, many measures have no Lebesgue density: $\delta_0$ is the simplest example.

For a nonnegative $\lambda$-measurable weight $w$, define formally

$$Z=\int_Xw\,\mathrm d\lambda.$$

The measure $w\,\mathrm d\lambda$ is generally unnormalized. It becomes the probability measure $Z^{-1}w\,\mathrm d\lambda$ only when $0<Z<\infty$. Positivity, measurability, and normalizability are separate checks.

A finite Euclidean Gaussian measure

Consider a real scalar field after restricting to $N<\infty$ independent real momentum-mode coordinates in a fixed orthonormal basis. Its configuration space is

$$(\mathbb R^N,\mathcal B(\mathbb R^N)).$$

Let $K$ be a real symmetric positive-definite $N\times N$ matrix and set

$$S_E(q)=\frac12q^{\mathsf T}Kq.$$

With ordinary Lebesgue base measure $\mathrm d^Nq$, the normalized Euclidean Gaussian measure is

$$\mathrm d\mu_K(q) = \frac{(\det K)^{1/2}}{(2\pi)^{N/2}} \exp\!\left(-\frac12q^{\mathsf T}Kq\right) \mathrm d^Nq.$$

This is a Borel probability measure. To check the normalization, write

$$K=O^{\mathsf T}\operatorname{diag}(\kappa_1,\ldots,\kappa_N)O, \qquad y=Oq, \qquad |\det O|=1.$$

All $\kappa_j$ are positive, $\mathrm d^Nq=\mathrm d^Ny$, and the resulting finite product of ordinary one-dimensional Gaussian integrals gives

$$\begin{aligned} Z_K &=\int_{\mathbb R^N} \exp\!\left(-\frac12q^{\mathsf T}Kq\right)\mathrm d^Nq\\ &=\prod_{j=1}^{N} \int_{-\infty}^{\infty} e^{-\kappa_jy_j^2/2}\,\mathrm dy_j\\ &=(2\pi)^{N/2}(\det K)^{-1/2}. \end{aligned}$$

This determinant normalization is the finite-dimensional Gaussian formula used in Zinn-Justin 2021, § 1.1.

For a fixed $v\in\mathbb R^N$, the mode observable

$$F_v(q)=v^{\mathsf T}q$$

is continuous and therefore Borel measurable. Every Borel question about its value, such as

$$F_v^{-1}([-R,R]) = \{q:|v^{\mathsf T}q|\leq R\},$$

is consequently a measurable event in configuration space. Its law is the pushforward $(F_v)_{\#}\mu_K$.

For $v\neq0$, set $y=K^{1/2}q$ and $a=K^{-1/2}v$. The $y$-density is the standard $N$-dimensional Gaussian, while $F_v(q)=a^{\mathsf T}y$. Choose an orthogonal matrix $R$ with $Ra=\lVert a\rVert e_1$ and put $z=Ry$. Orthogonal invariance leaves the standard Gaussian density and $\mathrm d^Ny$ unchanged, and

$$F_v(q)=\lVert a\rVert z_1, \qquad \lVert a\rVert^2=v^{\mathsf T}K^{-1}v.$$

The normalized factors in $z_2,\ldots,z_N$ integrate to one. The remaining one-dimensional factor therefore gives

$$\mathrm d\bigl((F_v)_{\#}\mu_K\bigr)(u) = \frac{1}{\sqrt{2\pi\sigma_v^2}} \exp\!\left(-\frac{u^2}{2\sigma_v^2}\right)\mathrm du, \qquad \sigma_v^2=v^{\mathsf T}K^{-1}v.$$

For $v=0$, the pushforward is instead $\delta_0$. This illustrates the precise chain

$$(\mathbb R^N,\mathcal B,\mu_K) \longrightarrow F_v \longrightarrow (F_v)_{\#}\mu_K:$$

configuration space, measurable observable, and output distribution are different objects.

Physics notation and mathematical status

Expression	Precise status
$\mathrm d^dp/(2\pi)^d$	scaled Lebesgue measure on continuous momentum-label space; sigma-finite, not a probability on all of $\mathbb R^d$
$e^{-S_E(q)}\mathrm d^Nq$	positive finite-dimensional measure when $S_E$ is real-valued and Borel measurable; generally unnormalized
$Z^{-1}e^{-S_E(q)}\mathrm d^Nq$	probability measure only when $0<Z<\infty$
$e^{iS(q)}\mathrm d^Nq$	oscillatory weight, not a positive probability density
$\mathcal D\phi$	formal continuum notation, not a measure constructed by the symbol alone

The positive-definiteness of $K$ is not cosmetic. A zero eigenvalue makes the integral over that direction diverge; a negative eigenvalue makes the Euclidean exponential grow. In either case the displayed normalization fails on $\mathbb R^N$.

This example proves only a finite-$N$, positive Euclidean statement. It does not construct an infinite-dimensional Lebesgue measure, justify a continuum limit, or turn a Lorentzian weight into a probability. The finite-mode and lattice construction, including the distinction between a defined regulator and formal continuum notation, continues in Regulated Bosonic Field Integrals. Its broader first application lies in Foundations, where momentum-space integration and probability measures enter regulated QFT calculations.

For the measure-theoretic distinctions used throughout this page, compare Folland 1999, §§ 1.1–1.5 and 2.1 and Tao 2011, §§ 1.2 and 1.4.2–1.4.4, PDF.

Common pitfalls

Invalid shortcut	Correction
Assign $\mu(A)$ before checking $A\in\Sigma$	A measure is defined only on its sigma-algebra.
Every subset of a measurable set is measurable	This is guaranteed only for subsets of measurable null sets in a complete space.
Borel measurable means Lebesgue measurable and conversely	Every Borel set is Lebesgue measurable, but completion adds non-Borel sets.
Null means empty or impossible	Nullity depends on the measure and does not imply topological smallness.
Almost everywhere means everywhere	The exceptional null set may be nonempty or even dense.
Direct images test measurability	Measurability is defined by inverse images of target measurable sets.
A density is the measure	A density is relative to a named reference measure; some measures have no such density.
Sigma-finite means normalized	Sigma-finite measures can have infinite total mass.
Any decreasing sequence is continuous from above	A finite-measure hypothesis is needed.
$e^{iS}\mathcal D\phi$ is already a probability measure	It is neither positive nor constructed merely by writing the formal symbol.

Where the next pages begin

• Lebesgue Integration and Convergence Theorems builds the integral from simple functions and proves monotone and dominated convergence.
• Product Measures, Fubini–Tonelli, and Change of Variables treats repeated integration, Jacobians, and transformations of densities.
• $L^p$ Spaces, Inequalities, and Weak Convergence identifies functions equal almost everywhere and develops norm and weak structures.
• Probability Spaces, Random Variables, and Conditional Expectation specializes normalized measure spaces to probability and develops expectation, independence, and conditioning.
• Regulated Bosonic Field Integrals develops the physical finite-mode construction and the continuum-measure warning.

Exercises

Generated sigma-algebra. Let $X=\{1,2,3,4\}$ and $\mathcal G=\{\{1,2\}\}$. Find $\sigma(\mathcal G)$.

Solution

Closure under complements forces $\{3,4\}$, and every sigma-algebra already contains $\varnothing$ and $X$. These four sets are closed under complements and countable unions, so

$$\sigma(\mathcal G) = \{\varnothing,\{1,2\},\{3,4\},X\}.$$

No singleton is forced by the generator.

Continuity hypothesis. For Lebesgue measure on $\mathbb R$, analyze $A_n=[n,\infty)$ and identify the failed hypothesis in continuity from above.

Solution

The sets decrease and their intersection is empty, but every $A_n$ has infinite measure. Hence

$$\lim_{n\to\infty}\lambda(A_n)=\infty \neq 0=\lambda\!\left(\bigcap_nA_n\right).$$

No member of the decreasing sequence has finite measure, so the subtraction argument from a finite containing set is unavailable.

Completion check. In the three-point example, why is $\mathbf1_{\{a\}}$ not measurable before completion, and why does it become measurable after completion?

Solution

Before completion,

$$\mathbf1_{\{a\}}^{-1}(\{1\})=\{a\}\notin\Sigma,$$

so the inverse-image test fails. The set $\{a\}$ is a subset of the measurable null set $\{a,b\}$. Completion adds it as a measurable null set, after which the indicator is measurable.

Pushforward check. Let $\mu$ be a finite measure, let $f:X\to Y$ be measurable, and let $f(x)=y_0$ for every $x$. Compute $f_{\#}\mu$.

Solution

For $B\in\mathcal T$, the inverse image is $X$ when $y_0\in B$ and empty otherwise. Therefore

$$f_{\#}\mu(B) = \begin{cases} \mu(X),&y_0\in B,\\ 0,&y_0\notin B, \end{cases}$$

so $f_{\#}\mu=\mu(X)\delta_{y_0}$. If $\mu$ is a probability measure, this is simply $\delta_{y_0}$.

QFT transfer. Classify each expression as a base measure, an unnormalized positive measure, a probability measure, or an expression that is not a positive probability measure:

$$\frac{\mathrm d^dp}{(2\pi)^d}, \qquad e^{-S_E(q)}\mathrm d^Nq, \qquad Z^{-1}e^{-S_E(q)}\mathrm d^Nq, \qquad e^{iS(q)}\mathcal Dq.$$

Solution

The first is a scaled Lebesgue base measure on momentum space. The second is a positive, generally unnormalized finite-dimensional measure when $S_E$ is real-valued and Borel measurable. The third is a probability measure only if $0<Z<\infty$. The Lorentzian expression is oscillatory rather than positive; the formal $\mathcal Dq$ symbol also supplies no measure construction by itself.

References

• Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Wiley (1999), publisher record, §§1.1–1.5 and 2.1. Structural reference for measures, outer measures, Borel measures, completion, and measurable functions.
• Andrew Lin, lecture notes for Casey Rodriguez’s 18.102 Introduction to Functional Analysis, Spring 2021, MIT OpenCourseWare, Lecture 7 PDF and Lecture 9 PDF. Sigma-algebras, Lebesgue measurable sets, real measurable functions, and pointwise-limit closure.
• Richard B. Melrose, Lecture Notes for 18.155, Fall 2004, MIT OpenCourseWare, official PDF, §§2–3. Generated and Borel sigma-algebras, measures, and measurable maps.
• Terence Tao, An Introduction to Measure Theory, Graduate Studies in Mathematics 126, American Mathematical Society (2011), author’s preliminary PDF, §§1.2 and 1.4.2–1.4.4. Lebesgue outer measure, sigma-algebras, measures, completeness, measurable maps, and pushforwards.
• Jean Zinn-Justin, Quantum Field Theory and Critical Phenomena, 5th ed., Oxford University Press (2021), DOI:10.1093/oso/9780198834625.003.0001, §1.1. Finite-dimensional positive Gaussian integrals, determinants, and normalization.