In this paper, we use Lebesgue measure to define the ${\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}$ of functions $f:R^d\to C\cup \left\{\infty \right\}$, where $R^d$ is the Euclidean space.

If $f=c_1{1}_{E_1}+\cdots +c_k{1}_{E_k}$ is an unsigned simple function, the integral ${\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}$ is defined by the formula

$Simp{\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}:=c_{1}m\left(E_1\right)+\cdots +c_{k}m\left(E_k\right)$, (1)

thus ${\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}$ will take values in $\left[0,+\infty \right]$.

Let $k,k^{\prime }\ge 0$ be natural numbers, $c_1,\cdots ,c_k,{c^{\prime }}_1,\cdots ,{c^{\prime }}_{k^{\prime }}\in \left[0,+\infty \right]$, and let $E_1,\cdots ,E_k,{E^{\prime }}_1,\cdots ,{E^{\prime }}_{k^{\prime }}\subset R^d$ be Lebesgue measurable sets such that the identity

$c_1{1}_{E_1}+\cdots +c_k{1}_{E_k}={c^{\prime }}_1{1}_{{E^{\prime }}_1}+\cdots +{c^{\prime }}_{k^{\prime }}{1}_{{E^{\prime }}_{k^{\prime }}}$ (2)

holds identically on $R^d$. Then one has

$c_{1}m\left(E_1\right)+\cdots +c_{k}m\left(E_k\right)={c^{\prime }}_{1}m\left({E^{\prime }}_1\right)+\cdots +{c^{\prime }}_{k^{\prime }}m\left({E^{\prime }}_{k^{\prime }}\right)$. (3)

A complex-valued simple function $f:R^d\to C$ is said to be absolutely integrable of ${\displaystyle {\int }_{R^d}\left|f\left(x\right)\right|\text{d}x}<\infty$. If $f$ is absolutely integrable, the integral ${\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}$ is defined for real signed $f$ by the formula

$Simp{\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}:=Simp{\displaystyle {\int }_{R^d}{f}_{+}\left(x\right)\text{d}x}-Simp{\displaystyle {\int }_{R^d}{f}_{-}\left(x\right)\text{d}x}$ (4)

where $f_{+}\left(x\right):=\text{max}\left(f\left(x\right),0\right)$ and $f_{-}\left(x\right):=\text{max}\left(-f\left(x\right),0\right)$ (we note that these are unsigned simple functions that are pointwise dominated by $\left|f\right|$ and thus have finite integral), and for complex-valued $f$ by the formula

$Simp{\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}:=Simp{\displaystyle {\int }_{R^d}\mathrm{Re}f\left(x\right)\text{d}x}+iSimp{\displaystyle {\int }_{R^d}\mathrm{Im}f\left(x\right)\text{d}x}$. (5)

Let $f:R^d\to \left[0,+\infty \right]$ be an unsigned function (not necessarily measurable). We define the lower unsigned Lebesgue integral ${\displaystyle {\int }_{\underset{\_}{R^d}}f\left(x\right)\text{d}x}$

${\displaystyle {\int }_{\underset{\_}{R^d}}f\left(x\right)\text{d}x}:={\sup }_{0\le g\le f;gsimple}Simp{\displaystyle {\int }_{\underset{\_}{R^d}}g\left(x\right)\text{d}x}$ (6)

where $g$ ranges over all unsigned simple functions $g:R^d\to \left[0,+\infty \right]$ that are pointwise bounded by $f$. One can also define the upper unsigned Lebesgue integral

${\displaystyle {\int }_{R^d}^{-}f\left(x\right)\text{d}x}:={\inf }_{h\ge f;hsimple}Simp{\displaystyle {\int }_{R^d}h\left(x\right)\text{d}x}$ (7)

but we will use this integral much more rarely. Note that both integrals take values in $\left[0,+\infty \right]$, and that the upper Lebesgue integral is always at least as large as the lower Lebesgue integral.

Let $f:R^d\to \left[0,+\infty \right]$ be measurable. Then for any $0<\lambda <\infty$, one has

$m\left(\left\{x\in R^d:f\left(x\right)\ge \lambda \right\}\right)\le \frac{1}{\lambda }{\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}$, (7b)

that is the Markov’s inequality.

An almost everywhere defined measurable function $f:R^d\to C$ is said be absolutely integrable if the unsigned integral

${\Vert f\Vert }_{L^1\left(R^d\right)}:={\displaystyle {\int }_{R^d}\left|f\left(x\right)\right|\text{d}x}$ (8)

is finite. We refer to this quantity ${\Vert f\Vert }_{L^1\left(R^d\right)}$ as the $L^1\left(R^d\right)$ norm of $f$, and use $L^1\left(R^d\right)$ or $L^1\left(R^d\to C\right)$ to denote the space of absolutely integrable functions. If $f$ is real-valued and absolutely integrable, we define the Lebesgue integral ${\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}$ by the formula

${\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}:={\displaystyle {\int }_{R^d}{f}_{+}\left(x\right)\text{d}x}-{\displaystyle {\int }_{R^d}{f}_{-}\left(x\right)\text{d}x}$ (9)

where $f_{+}:=\text{max}\left(f,0\right)$ and $f_{-}:=\text{max}\left(-f,0\right)$ are the positive and negative components of $f$. If $f$ is complex-valued and absolutely integrable, we define the Lebesgue integral ${\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}$ by the formula

${\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}:={\displaystyle {\int }_{R^d}\mathrm{Re}f\left(x\right)\text{d}x}+i{\displaystyle {\int }_{R^d}\mathrm{Im}f\left(x\right)\text{d}x}$ (10)

where the two integrals on the right are interpreted as real-valued absolutely integrable Lebesgue integrals.

Let $f\in L^1\left(R^d\to C\right)$. Then

$\left|{\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}\right|\le {\displaystyle {\int }_{R^d}\left|f\left(x\right)\right|\text{d}x}$. (11)

This is the Triangle inequality.

If $f$ is real-valued, then $\left|f\right|=f_{+}+f_{-}$. When $f$ is complex-value one cannot argue quite so simply; a naive mimicking of the real-valued argument would lose a factor of 2, giving the inferior bound

$\left|{\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}\right|\le 2{\displaystyle {\int }_{R^d}\left|f\left(x\right)\right|\text{d}x}$. (12)

To do better, we exploit the phase rotation invariance properties of the absolute value operation and of the integral, as follows. Note that for any complex number $z$, one can write $\left|z\right|$ as $z{\text{e}}^{i\theta }$ for some real $\theta$. In particular, we have

$\left|{\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}\right|={\text{e}}^{i\theta }{\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}={\displaystyle {\int }_{R^d}{\text{e}}^{i\theta }f\left(x\right)\text{d}x}$, (13)

for some real $\theta$. Taking real parts of both sides, we obtain

$\left|{\displaystyle {\int }_{R^d}f\left(x\right)\text{d}x}\right|={\displaystyle {\int }_{R^d}\text{Re}\left({\text{e}}^{i\theta }f\left(x\right)\right)\text{d}x}$. (14)

Since

$\text{Re}\left({\text{e}}^{i\theta }f\left(x\right)\right)\le \left|{\text{e}}^{i\theta }f\left(x\right)\right|=\left|f\left(x\right)\right|$,

we obtain the Equation (11)

Let $\left(X,B,\mu \right)$ be a measure space, and let $f,g:X\to \left[0,+\infty \right]$ be measurable. Then

${\displaystyle {\int }_X\left(f+g\right)\text{d}\mu }={\displaystyle {\int }_{X}f\text{d}\mu }+{\displaystyle {\int }_{X}g\text{d}\mu }$. (15)

It suffices to establish the sub-additivity property

${\displaystyle {\int }_X\left(f+g\right)\text{d}\mu }\le {\displaystyle {\int }_{X}f\text{d}\mu }+{\displaystyle {\int }_{X}g\text{d}\mu }$. (15b)

We establish this in stages. We first deal with the case when $\mu$ is a finite measure (which means that $\mu \left(X\right)<\infty$) and $f,g$ are bounded. Pick an $ϵ>0$ and $f_{ϵ}$ be $f$ rounded down to the nearest integer multiple of $ϵ$, and $f^{ϵ}$ be $f$ rounded up to nearest integer multiple. Clearly, we have the pointwise bound

$f_{ϵ}\left(x\right)\le f\left(x\right)\le f^{ϵ}\left(x\right)$ (16)

and

$f^{ϵ}\left(x\right)-f_{ϵ}\left(x\right)\le ϵ$. (17)

Since $f$ is bounded, $f_{ϵ}$ and $f^{ϵ}$ are simple. Similarly define $g_{ϵ},g^{ϵ}$. We then have the pointwise bound

$f+g\le f^{ϵ}+g^{ϵ}\le f_{ϵ}+g_{ϵ}+2ϵ$, (18)

hence, from the properties of the simple integral,

$\begin{array}{c}{\displaystyle {\int }_{X}f+g\text{d}\mu }\le {\displaystyle {\int }_X{f}_{ϵ}+g_{ϵ}+2ϵ\text{d}\mu }=Simp{\displaystyle {\int }_X{f}_{ϵ}+g_{ϵ}+2ϵ\text{d}\mu }\\ =Simp{\displaystyle {\int }_X{f}_{ϵ}\text{d}\mu }+Simp{\displaystyle {\int }_X{g}_{ϵ}\text{d}\mu }+2ϵ\mu \left(X\right).\end{array}$ (19)

From the following equation

${\displaystyle {\int }_{X}f\text{d}\mu }:={\sup }_{0\le g\le f;gsimple}Simp{\displaystyle {\int }_{X}g\text{d}\mu }$. (19b)

we conclude that

${\displaystyle {\int }_{X}f+g\text{d}\mu }\le {\displaystyle {\int }_{X}f\text{d}\mu }+{\displaystyle {\int }_{X}g\text{d}\mu }+2ϵ\mu \left(X\right)$. (20)

Letting $ϵ\to 0$ and using the assumption that $\mu \left(X\right)$ is finite, we obtain the claim. Now we continue to assume that $\mu$ is a finite measure, but now do not assume that $f,g$ are bounded. Then for any natural number (also the primes) $n$, we can use the previous case to deduce that

${\displaystyle {\int }_X\text{min}\left(f,n\right)+\text{min}\left(g,n\right)\text{d}\mu }\le {\displaystyle {\int }_X\text{min}\left(f,n\right)\text{d}\mu }+{\displaystyle {\int }_X\text{min}\left(g,n\right)\text{d}\mu }$. (21)

Since $\text{min}\left(f+g,n\right)\le \text{min}\left(f,n\right)+\text{min}\left(g,n\right)$, we conclude that

${\displaystyle {\int }_X\text{min}\left(f+g,n\right)}\le {\displaystyle {\int }_X\text{min}\left(f,n\right)\text{d}\mu }+{\displaystyle {\int }_X\text{min}\left(g,n\right)\text{d}\mu }$. (22)

Taking limits as $n\to \infty$ using horizontal truncation, we obtain the claim.

Finally, we no longer assume that $\mu$ is of finite measure, and also do not require $f,g$ to be bounded. By Markov’s inequality, we see that for each natural number (also the primes) $n$, the set

$E_n:=\left\{x\in X:f\left(x\right)>\frac{1}{n}\right\}\cup \left\{x\in X:g\left(x\right)>\frac{1}{n}\right\}$,

has finite measure. These sets are increasing in $n$, and $f,g,f+g$ are supported on ${\displaystyle {\cup }_{n=1}^{\infty }{E}_n}$, and so by vertical truncation

${\displaystyle {\int }_X\left(f+g\right)\text{d}\mu }={\lim }_{n\to \infty }{\displaystyle {\int }_X\left(f+g\right)1_{E_n}\text{d}\mu }$. (23)

From the previous case, we have

${\displaystyle {\int }_X\left(f+g\right)1_{E_n}\text{d}\mu }\le {\displaystyle {\int }_{X}f{1}_{E_n}\text{d}\mu }+{\displaystyle {\int }_{X}g{1}_{E_n}\text{d}\mu }$. (24)

Let $\left(X,B,\mu \right)$ be a measure space, and let $0\le f_1\le f_2\le \cdots$ be a monotone non-decreasing sequence of unsigned measurable functions on $X$. Then we have

. (25)

Write $f:={\lim }_{n\to \infty }{f}_n={\sup }_{n\to \infty }{f}_n$, then $f:X\to \left[0,+\infty \right]$ is measurable. Since the $f_n$ are non decreasing to $f$, we see from monotonicity that ${\displaystyle {\int }_X{f}_n\text{d}\mu }$ are non decreasing and bounded above by ${\displaystyle {\int }_{X}f\text{d}\mu }$, which gives the bound

${\lim }_{n\to \infty }{\displaystyle {\int }_X{f}_n\text{d}\mu }\le {\displaystyle {\int }_{X}f\text{d}\mu }$. (26)

It remains to establish the reverse inequality

${\displaystyle {\int }_{X}f\text{d}\mu }\le {\lim }_{n\to \infty }{\displaystyle {\int }_X{f}_n\text{d}\mu }.$. (27)

By definition, it suffices to show that

${\displaystyle {\int }_{X}g\text{d}\mu }\le {\lim }_{n\to \infty }{\displaystyle {\int }_X{f}_n\text{d}\mu }$, (28)

whenever $g$ is a simple function that is bounded pointwise by $f$. By horizontal truncation we may assume without loss of generality that $g$ also is finite everywhere, then we can write

$g={\displaystyle {\sum }_{i=1}^k{c}_i{1}_{A_i}}$ (29)

for some $0\le c_i<\infty$ and some disjoint $B$-measurable sets $A_1,\cdots ,A_k$, thus

${\displaystyle {\int }_{X}g\text{d}\mu }={\displaystyle {\sum }_{i=1}^k{c}_i\mu \left(A_i\right)}$. (30)

Let $0<ϵ<1$ be arbitrary (also $\frac{\sqrt{5}-1}{2}=0,61803398\cdots$). Then we have

(31)

for all $x\in A_i$. Thus, if we define the sets

$A_{i,n}:=\left\{x\in A_i:f_n\left(x\right)>\left(1-ϵ\right)c_i\right\}$ (32)

then the $A_{i,n}$ increase to $A_i$ and are measurable. By upwards monotonicity of measure, we conclude that

${\lim }_{n\to \infty }\mu \left(A_{i,n}\right)=\mu \left(A_i\right)$ (33)

On the other hand, observe the pointwise bound

$f_n\ge {\displaystyle {\sum }_{i=1}^k\left(1-ϵ\right)c_i{1}_{A_{i,n}}}$ (34)

for any $n$; integrating this, we obtain

${\displaystyle {\int }_X{f}_n\text{d}\mu }\ge \left(1-ϵ\right){\displaystyle {\sum }_{i=1}^k{c}_i\mu \left(A_{i,n}\right)}$. (35)

Taking limits as $n\to \infty$, we obtain

${\lim }_{n\to \infty }{\displaystyle {\int }_X{f}_n\text{d}\mu }\ge \left(1-ϵ\right){\displaystyle {\sum }_{i=1}^k{c}_i\mu \left(A_i\right)}$, (36)

sending $ϵ\to 0$ we then obtain the claim.

Let $\left(X,B,\mu \right)$ be a measure space, and let $f_1,f_2,\cdots :X\to C$ be a sequence of measurable functions that converge pointwise $\mu$-almost everywhere to a measurable limit $f:X\to C$. Suppose that there is an unsigned absolutely integrable function $G:X\to \left[0,+\infty \right]$ such that $\left|f_n\right|$ are pointwise $\mu$-almost everywhere bounded by $G$ for each $n$. Then we have

${\lim }_{n\to \infty }{\displaystyle {\int }_X{f}_n\text{d}\mu }={\displaystyle {\int }_{X}f\text{d}\mu }$. (37)

By modifying $f_n,f$ on a null set, we may assume without loss of generality that the $f_n$ converge to $f$ pointwise everywhere rather than $\mu$-almost everywhere, and similarly we can assume that $\left|f_n\right|$ are bounded by $G$ pointwise everywhere rather than $\mu$-almost everywhere. By taking real and imaginary parts we may assume without loss of generality that $f_n,f$ are real, thus $-G\le f_n\le G$ pointwise. Of course, this implies that $-G\le f\le G$ pointwise also. If we apply Fatou’s lemma to the unsigned functions $f_n+G$, we see that

${\displaystyle {\int }_{X}f+G\text{d}\mu }\le {\lim }_{n\to \infty }\text{inf}{\displaystyle {\int }_X{f}_n+G\text{d}\mu }$, (38)

which on subtracting the finite quantity ${\displaystyle {\int }_{X}G\text{d}\mu }$ gives

${\displaystyle {\int }_{X}f\text{d}\mu }\le {\lim }_{n\to \infty }\text{inf}{\displaystyle {\int }_X{f}_n\text{d}\mu }$. (39)

Similarly, if we apply that lemma to the unsigned functions $G-f_n$, we obtain

${\displaystyle {\int }_{X}G-f\text{d}\mu }\le {\lim }_{n\to \infty }\text{inf}{\displaystyle {\int }_{X}G-f_n\text{d}\mu }$; (40)

negating this inequality and then cancelling ${\displaystyle {\int }_{X}G\text{d}\mu }$ again we conclude that

${\lim }_{n\to \infty }\text{sup}{\displaystyle {\int }_X{f}_n\text{d}\mu }\le {\displaystyle {\int }_{X}f\text{d}\mu }$. (41)

The claim then follows by combining these inequalities.

A probability space is a measure space $\left(\text{Ω},F,P\right)$ of total measure 1: $P\left(\text{Ω}\right)=1$. The measure $P$ is known as a probability measure. If $\text{Ω}$ is a (possibility infinite) non-empty set with the discrete $\sigma$-algebra $2^{\text{Ω}}$, and if ${\left(p_{\omega }\right)}_{\omega \in \text{Ω}}$ are a collection of real numbers in $\left[0,1\right]$ with ${\displaystyle {\sum }_{\omega \in \text{Ω}}{p}_{\omega }}=1$, then the probability measure $P$ defined by $P:={\displaystyle {\sum }_{\omega \in \text{Ω}}{p}_{\omega }{\delta }_{\omega }}$, or in other words

$P\left(E\right):=\underset{\omega \in E}{{\displaystyle \sum }}{p}_{\omega }$, (42)

is indeed a probability measure, and $\left(\text{Ω},2^{\text{Ω}},P\right)$ is a probability space. The function $\omega \mapsto p_{\omega }$ is known as the (discrete) probability distribution of the state variable $\omega$. Similarly, if $\text{Ω}$ is a Lebesgue measurable subset of $R^d$ of positive (and possibly infinite) measure, and $f:\text{Ω}\to \left[0,+\infty \right]$ is a Lebesgue measurable function on $\text{Ω}$ (where of course we restrict the Lebesgue measure space on $R^d$

to $\text{Ω}$ in the usual fashion) with ${\displaystyle {\int }_{\text{Ω}}f\left(x\right)\text{d}x}=1$, then $\left(\text{Ω},ℒ\left[R^d\right]{\downarrow }_{\text{Ω}},P\right)$ is a

probability space, where $P:=m_f$ is the measure

$P\left(E\right):={\displaystyle {\int }_{\text{Ω}}{1}_E\left(x\right)f\left(x\right)\text{d}x}={\displaystyle {\int }_{E}f\left(x\right)\text{d}x}$. (43)

The function $f$ is known as the (continuous) probability density of the state variable $\omega$.

Theorem 1

(Connes’ Trace Theorem) Let $M$ be a compact $n$-dimensional manifold, $\xi$ a complex vector bundle on $M$, and $P$ a pseudo-differential operator of order- $n$ acting on sections of $\xi$. Then the corresponding operator $P$ in $H=L^2\left(M,\xi \right)$ belongs to ${ℒ}^{1,\infty }\left(H\right)$ and one has:

$T{r}_{\omega }\left(P\right)=\frac{1}{n}\text{Res}\left(P\right)$ (44)

for any $\omega$.

Here Res is the restriction of the Adler-Manin-Wodzicki residue to pseudo-differential operators of order- $n$. Let $\xi$ be the exterior bundle on a (closed) compact Riemannian manifold $M$, $\left|vol\right|$ the 1-density of $M$, $f\in C^{\infty }\left(M\right)$, $M_f$ the operator given by $f$ acting by multiplication on smooth sections of $\xi$, $\text{Δ}$ the Hodge Laplacian on smooth sections of $\xi$, and $P=M_f{\left(1+\text{Δ}\right)}^{-n/2}$, which is a pseudo-differential operator of order- $n$. Using Theorem 1, we have that

${\varphi }_{\omega }\left(M_f\right)=T{r}_{\omega }\left(M_f{T}_{\text{Δ}}\right)=\frac{1}{2^{n-1}{\pi }^{\frac{n}{2}}\text{Γ}\left(\frac{n}{2}+1\right)}{\displaystyle {\int }_{M}f}\left(x\right)\left|vol\right|\left(x\right),f\in C^{\infty }\left(M\right)$ (45)

where we set $T_{\text{Δ}}:={\left(1+\text{Δ}\right)}^{-n/2}\in {ℒ}^{1,\infty }$. This has become the standard way to identify ${\varphi }_{\omega }$ with the Lebesgue integral for $f\in C^{\infty }\left(M\right)$.

Corollary 1

Let $M$ be a $n$-dimensional (closed) compact Riemannian manifold with Hodge Laplacian $\text{Δ}$. Set $T_{\text{Δ}}:={\left(1+\text{Δ}\right)}^{-n/2}\in {ℒ}^1{}^{,\infty }\left(L^2\left(M\right)\right)$. Then

${\varphi }_{\omega }\left(M_f\right):=T{r}_{\omega }\left(M_f{T}_{\text{Δ}}\right)=c{\displaystyle {\int }_{M}f}\left(x\right)\left|vol\right|\left(x\right),\forall f\in L^{\infty }\left(M\right)$ (46)

where $c>0$ is a constant independent of $\omega \in D{L}_2$.

Theorem 2

Let $M,\text{Δ},T_{\text{Δ}}$ be as in Corollary 1. Then, ${\langle M_f\rangle }_{T_s^s}=T_{\text{Δ}}^{s/2}{M}_f{T}_{\text{Δ}}^{s/2}\in {ℒ}^1\left(L^2\left(M\right)\right)$ for all $s>1$ if and only if $f\in L^1\left(M\right)$. Moreover, setting

${\psi }_{\xi }\left(M_f\right):=\xi \left(\frac{1}{k}Tr\left({\langle M_f\rangle }_{T_{\text{Δ}}^{1+\frac{1}{k}}}\right)\right)$ (47)

for any $\xi \in BL$,

${\psi }_{\xi }\left(M_f\right):=\underset{k\to \infty }{\lim }\frac{1}{k}Tr\left({\langle M_f\rangle }_{T_{\text{Δ}}^{1+\frac{1}{k}}}\right)=c{\displaystyle {\int }_{M}f}\left(x\right)\left|vol\right|\left(x\right),\forall f\in L^1\left(M\right)$ (48)

for a constant $c>0$ independent of $\xi \in BL$.

Thus ${\psi }_{\xi }$, as the residue of the zeta function $Tr\left(T_{\text{Δ}}^{s/2}{M}_f{T}_{\text{Δ}}^{s/2}\right)$ at $s=1$, is the value of the Lebesgue integral of the integrable function $f$ on $M$. This is the most general form of the identification between the Lebesgue integral and an algebraic expression involving $M_f$, the compact operator ${\left(1+{\text{Δ}}^2\right)}^{-1}$ and a trace.

Theorem 3

Let $0<G\left(D\right)\in {ℒ}^{1,\infty }$ and $\xi \in BL\cap DL$. Then

${\varphi }_{ℒ\left(\xi \right)}\left(T_f\right):=T{r}_{ℒ\left(\xi \right)}\left(T_{f}G\left(D\right)\right)=\xi \left(\frac{1}{k}{\displaystyle {\int }_{F}f}\left(x\right)\text{d}{\mu }_{1+\frac{1}{k}}\left(x\right)\right),\forall f\in L_0^2\left(F,{\mu }_{1,\infty }\right)$. (49)

Moreover, if ${\lim }_{k\to \infty }{k}^{-1}{\displaystyle {\int }_{F}h}\left(x\right)\text{d}{\mu }_{1+k^{-1}}\left(x\right)$ exists for all $h\in L^{\infty }\left(F,{\mu }_{1,\infty }\right)$, then

${\varphi }_{\omega }\left(T_f\right):=T{r}_{\omega }\left(T_{f}G\left(D\right)\right)={\lim }_{k\to \infty }\frac{1}{k}{\displaystyle {\int }_{F}f}\left(x\right)\text{d}{\mu }_{1+\frac{1}{k}}\left(x\right),\forall f\in L_0^2\left(F,{\mu }_{1,\infty }\right)$ (50)

and all $\omega \in D{L}_2$.

We note that it is possible to identify ${\varphi }_{\omega }$ with the Lebesgue integral.

Now we consider an arbitrary manifold $X$ with a fixed continuous non-negative finite Borel measure $m$. The construction of the integral models of representations of the current groups $G^X$ is based on the existence, in the space $D\left(X\right)$ of Schwartz distributions on $X$, of a certain measure $ℒ$ which is an infinite-dimensional analogue of the Lebesgue measure. Furthermore, we have that $\xi$ runs over the points of the cone

$l_{+}^1\left(X\right)=\left\{\xi =\underset{k=1}{\overset{\infty }{{\displaystyle \sum }}}{r}_k{\delta }_{x_k}|r_k>0,\underset{k}{{\displaystyle \sum }}{r}_k<\infty ,x_k\in X\right\},$

on which the infinite-dimensional Lebesgue measure $ℒ$ is concentrated. With each finite partition of $X$ into measurable sets,

$\alpha :X={\displaystyle \underset{k=1}{\overset{n}{\cup }}{X}_k},m\left(X_k\right)={\lambda }_k,k=1,\cdots ,n,$

we associate the cone ${ℱ}_a=R_{+}^n$ of piecewise constant positive functions of the form

$f\left(x\right)=\underset{k=1}{\overset{n}{{\displaystyle \sum }}}{f}_k{\chi }_k\left(x\right),f_k>0$,

where ${\chi }_k$ is the characteristic function of $X_k$, and we denote by ${\text{Φ}}_a=\left(R_{+}^n\right)\text{'}$ the dual cone in the space distributions. We define a measure ${ℒ}_a$ on ${\text{Φ}}_a$ by

$\text{d}{ℒ}_a\left(x_1,\cdots ,x_n\right)=\underset{k=1}{\overset{n}{{\displaystyle \prod }}}\frac{x_k^{{\lambda }_k-1}\text{d}{x}_k}{\gamma \left({\lambda }_k\right)},\text{where}{\lambda }_k=m\left(X_k\right)$. (51)

Let $D_{+}\left(X\right)\subset D\left(X\right)$ be the set (cone) of non-negative Schwartz distributions on $X$, and let $l_{+}^1\left(X\right)\subset D_{+}\left(X\right)$ be the subset (cone) of discrete finite (non-negative) measures on $X$, that is,

$\phi \left(0,x\right)\equiv 1\text{and}{\displaystyle {\int }_X{\displaystyle {\int }_0^{\infty }\left(\phi \left(r,x\right)-{\text{e}}^{-r}\right)r^{-1}\text{d}r\text{d}m\left(x\right)}}<\infty$. (54)

Theorem

The following equality holds:

. (55)

Proof. Under the projection $D_{+}\left(X\right)\to {\text{Φ}}_a$ (recall that ${\text{Φ}}_a$ is the finite-dimensional space associated with a partition $\alpha :X={\displaystyle {\cup }_{k=1}^n{X}_k}$) the left-hand side of (53) takes the form

$I_a=\underset{k=1}{\overset{n}{{\displaystyle \prod }}}{I}_a^k,I_a^k=\frac{1}{\text{Γ}\left({\lambda }_k\right)}{\displaystyle {\int }_0^{\infty }{\phi }_{a,k}\left(r_k\right)r_k^{{\lambda }_{k-1}}\text{d}{r}_k}$, (56)

where ${\lambda }_k=m\left(X_k\right)$ and ${\phi }_{a,k}\left(r_k\right)={\lambda }_k^{-1}{\displaystyle {\int }_{X_k}\phi \left(r_k,x\right)\text{d}m\left(x\right)}$. Thence the Equation

(56) can be rewritten also as follows

$I_a=\underset{k=1}{\overset{n}{{\displaystyle \prod }}}\frac{1}{\text{Γ}\left({\lambda }_k\right)}{\displaystyle {\int }_0^{\infty }{\lambda }_k^{-1}}{\displaystyle {\int }_{X_k}\phi \left(r_k,x\right)\text{d}m\left(x\right)r_k^{{\lambda }_{k-1}}\text{d}{r}_k}$. (56b)

The original integral $I$ is the inductive limit of the integrals $I_a$ over the set

of partitions $\alpha$. Since $\frac{1}{\text{Γ}\left({\lambda }_k\right)}{\displaystyle {\int }_0^{\infty }{\text{e}}^{-t}{r}_k^{{\lambda }_k-1}\text{d}{r}_k}=1$, the integral $I_{\alpha }^k$ can be written

in the form

$I_a^k=1+\frac{1}{\text{Γ}\left({\lambda }_k\right)}{\displaystyle {\int }_0^{\infty }\left({\phi }_{\alpha ,k}\left(r_k\right)-{\text{e}}^{-r_k}\right)r_k^{{\lambda }_k-1}\text{d}{r}_k}$. (57)

It follows that

$I_a^k=1+{\lambda }_k{\displaystyle {\int }_0^{\infty }\left({\phi }_{\alpha ,k}\left(r_k\right)-{\text{e}}^{-r_k}\right)r_k^{-1}\text{d}{r}_k}+O\left({\lambda }_k^2\right)$, (58)

whence

$I_a^k=\text{exp}\left({\lambda }_k{\displaystyle {\int }_0^{\infty }\left({\phi }_{a,k}\left(r_k\right)-{\text{e}}^{-r_k}\right)r_k^{-1}\text{d}{r}_k}\right)+O\left({\lambda }_k^2\right)$.

The Equation (58) can be rewritten also as follows

$\begin{array}{c}{I}_a^k=1+{\lambda }_k{\displaystyle {\int }_0^{\infty }\left({\phi }_{a,k}\left(r_k\right)-{\text{e}}^{-r_k}\right)r_k^{-1}\text{d}{r}_k}+O\left({\lambda }_k^2\right)\\ =\text{exp}\left({\lambda }_k{\displaystyle {\int }_0^{\infty }\left({\phi }_{a,k}\left(r_k\right)-{\text{e}}^{-r_k}\right)r_k^{-1}\text{d}{r}_k}\right)+O\left({\lambda }_k^2\right).\end{array}$ (58b)

Thus, up to terms of order greater than 1 with respect to ${\lambda }_k$,

$I_a\cong \text{exp}\left(\underset{k=1}{\overset{n}{{\displaystyle \sum }}}{\lambda }_k{\displaystyle {\int }_0^{\infty }\left({\phi }_{a,k}\left(r\right)-{\text{e}}^{-r}\right)r^{-1}\text{d}r}\right)$

Since

${\displaystyle {\sum }_{k=1}^n\left({\lambda }_k\left({\phi }_{a,k}\left(r\right)-{\text{e}}^{-r}\right)\right)}={\displaystyle {\int }_X\left(\phi \left(r,x\right)-{\text{e}}^{-r}\right)\text{d}m}\left(x\right),$

the expression obtained can be written in the following form

$I_{\alpha }\cong \text{exp}\left({\displaystyle {\int }_X{\displaystyle {\int }_0^{\infty }\left(\phi \left(r,x\right)-{\text{e}}^{-r}\right)r^{-1}\text{d}r\text{d}m\left(x\right)}}\right)$. (59)

The proof is completed by taking the inductive limit over the set of partitions $\alpha$.

Corollary

If $\phi \left(r,x\right)={\displaystyle {\sum }_{i=1}^n{c}_i{\phi }_i\left(r,x\right)}$, where $c_i>0$, ${\displaystyle \sum c_i}=1$, and the functions ${\phi }_i\left(r,x\right)$ satisfy (54), then

. (60)

Let $\phi \left(r,x\right)={\text{e}}^{-r^{\sigma }a\left(x\right)}$, where $\sigma \ge 1$ and $\mathrm{Re}a\left(x\right)>0$. In this case we obtain

(61)

Let us integrate with respect to $r$. We have

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }\left({\text{e}}^{-r^{\sigma }a\left(x\right)}-{\text{e}}^{-r}\right)r^{-1}\text{d}r}=\underset{\lambda \to 0}{\text{lim}}\left({\displaystyle {\int }_0^{\infty }\left({\text{e}}^{-r^{\sigma }a\left(x\right)}-{\text{e}}^{-r}\right)r^{\lambda -1}\text{d}r}\right)\\ =\underset{\lambda \to 0}{\text{lim}}\left({\sigma }^{-1}\text{Γ}\left(\frac{\lambda }{\sigma }\right)a^{-\lambda /\sigma }\left(x\right)-\text{Γ}\left(\lambda \right)\right).\end{array}$ (62)

Since $\text{Γ}\left(\lambda \right)\approx {\lambda }^{-1}+\gamma$ as $\lambda \to 0$, where $\gamma$ is the Euler constant, it follows that

${\displaystyle {\int }_0^{\infty }\left(\text{exp}\left(-r^{\sigma }a\left(x\right)\right)-\text{exp}\left(-r\right)\right)r^{-1}\text{d}r}=-{\sigma }^{-1}\log a\left(x\right)+\left({\sigma }^{-1}-1\right)\gamma$.

Hence,

$\begin{array}{l}{\displaystyle {\int }_{l_{+}^1\left(X\right)}\left(\underset{k=1}{\overset{\infty }{{\displaystyle \prod }}}\exp \left(-r_k^{\sigma }a\left(x_k\right)\right)\right)\text{d}ℒ\left(\xi \right)}\\ =\exp \left(\left({\sigma }^{-1}-1\right)\gamma \right)\exp \left(-{\sigma }^{-1}{\displaystyle {\int }_X\log a\left(x\right)\text{d}m\left(x\right)}\right).\end{array}$ (63)

In particular, for $\sigma =1$ we recover the original formula for the Laplace transform of the measure $ℒ$:

${\displaystyle {\int }_{l_{+}^1\left(X\right)}\underset{k=1}{\overset{\infty }{{\displaystyle \prod }}}\exp \left(-{\displaystyle \sum r_{k}a\left(x_k\right)}\right)\text{d}ℒ\left(\xi \right)}=\exp \left(-{\displaystyle {\int }_X\log a\left(x\right)\text{d}m\left(x\right)}\right)$. (63b)