Lebesgue Integration and Convergence Theorems

For nonnegative measurable functions that increase almost everywhere, the monotone convergence theorem permits a limit to pass through the integral, even when the answer is $+\infty$. For almost-everywhere convergence under one fixed integrable absolute bound, the dominated convergence theorem gives the stronger conclusion

$$\int_X |f_n-f|\,\mathrm d\mu\longrightarrow 0,$$

and therefore convergence of the integrals. Fatou’s lemma is the one-sided fallback for an arbitrary nonnegative sequence. Pointwise or almost-everywhere convergence by itself licenses none of these conclusions.

The construction and all three theorems concern one fixed measure space $(X,\Sigma,\mu)$. The final application removes a field-amplitude cutoff from a finite-dimensional positive Euclidean Gaussian. Product measures, iterated integrals, the full theory of $L^p$ spaces, and continuum field measures are treated separately.

Required background. Measures and Measurable Functions supplies sigma-algebras, null sets, measurable functions, and approximation from below by simple functions.

Helpful background. Limits, Completeness, and Modes of Convergence develops the general discipline for interchanging limits with other operations.

Integral construction · Monotone convergence · Fatou’s lemma · Dominated convergence · Theorem selection · Convergence in measure · Radon–Nikodym · Gaussian cutoff · Exercises

From simple functions to the Lebesgue integral

Lebesgue integration begins with nonnegative simple functions and extends them by monotone approximation. The measure need not be Lebesgue measure.

Let

$$s=\sum_{j=1}^{m}a_j\mathbf 1_{A_j}, \qquad a_j\geq 0,$$

where the measurable sets $A_j$ are pairwise disjoint and $s=0$ outside their union. Define

$$\int_X s\,\mathrm d\mu = \sum_{j=1}^{m}a_j\mu(A_j),$$

using the convention $0\cdot\infty=0$. Splitting overlapping sets into disjoint level sets shows that the value does not depend on the chosen simple-function representation.

For a measurable function $f:X\to[0,\infty]$, define

$$\int_X f\,\mathrm d\mu = \sup\left\{ \int_X s\,\mathrm d\mu: 0\leq s\leq f,\ s\ \text{is a nonnegative simple function} \right\}.$$

The supremum is taken in $[0,\infty]$. Thus a nonnegative measurable function always has an integral in the extended sense; the value need not be finite. If $A\in\Sigma$, write

$$\int_A f\,\mathrm d\mu \equiv \int_X \mathbf 1_A f\,\mathrm d\mu.$$

This construction from simple functions and monotone approximation is developed in Lin 2021, Lecture 10, pp. 47–52, PDF.

The definition gives the following basic rules. For nonnegative measurable functions, all equalities are understood in $[0,\infty]$:

Rule	Statement
Monotonicity	$0\leq f\leq g$ a.e. implies $\int f\,\mathrm d\mu\leq\int g\,\mathrm d\mu$
Positive homogeneity	$\int cf\,\mathrm d\mu=c\int f\,\mathrm d\mu$ for $c\geq0$
Additivity	$\int(f+g)\,\mathrm d\mu=\int f\,\mathrm d\mu+\int g\,\mathrm d\mu$
Restriction	$\int_A f\,\mathrm d\mu=\int\mathbf1_Af\,\mathrm d\mu$
Null-set invariance	$f=g$ a.e. implies $\int f\,\mathrm d\mu=\int g\,\mathrm d\mu$

Positive homogeneity follows directly from the supremum definition. Additivity is first checked for simple functions and then obtained by simultaneously approximating both summands from below. Null-set invariance follows because a nonnegative function supported on a measurable null set has integral zero.

Signed and complex integrals

For a measurable real-valued function $f$, set

$$f^+=\max(f,0), \qquad f^-=\max(-f,0).$$

Then

$$f=f^+-f^-, \qquad |f|=f^++f^-.$$

The extended integral

$$\int_X f\,\mathrm d\mu = \int_X f^+\,\mathrm d\mu - \int_X f^-\,\mathrm d\mu$$

is defined when the right side is not the indeterminate expression $\infty-\infty$. The function is integrable precisely when

$$\int_X|f|\,\mathrm d\mu<\infty,$$

equivalently when both $\int f^+\,\mathrm d\mu$ and $\int f^-\,\mathrm d\mu$ are finite. The notation $L^1(\mu)$ will mean the integrable functions, with functions equal almost everywhere identified; the full normed-space theory begins on the later $L^p$ page.

For a measurable complex-valued function $f=u+iv$, define

$$\int_Xf\,\mathrm d\mu = \int_Xu\,\mathrm d\mu + i\int_Xv\,\mathrm d\mu$$

when $f$ is integrable. On $L^1(\mu)$, the integral is linear and satisfies

$$\left|\int_Xf\,\mathrm d\mu\right| \leq \int_X|f|\,\mathrm d\mu.$$

The qualifier “on $L^1$” matters: unrestricted algebra with extended signed integrals can manufacture the undefined expression $\infty-\infty$. The signed and complex extensions are treated in Lin 2021, Lecture 12, pp. 59–64, PDF and Folland 1999, §§2.2–2.3, pp. 49–59.

Monotone convergence

Write $f_n\uparrow f$ almost everywhere when there is one measurable null set outside which

$$0\leq f_1(x)\leq f_2(x)\leq\cdots \qquad\text{and}\qquad \lim_{n\to\infty}f_n(x)=f(x).$$

Monotone convergence theorem. Let $f_n:X\to[0,\infty]$ and $f:X\to[0,\infty]$ be measurable. If $f_n\uparrow f$ almost everywhere, then

$$\int_Xf_n\,\mathrm d\mu \uparrow \int_Xf\,\mathrm d\mu.$$

The common limiting value may be $+\infty$. No assumption that $\mu(X)<\infty$ or that any integral is finite is required. Parallel statements and proofs appear in Lin 2021, Lecture 11, pp. 53–55, PDF and Tao 2011, Theorem 1.4.44, pp. 107–108, author-hosted preliminary PDF.

Proof

The exceptional sets for monotonicity and convergence have a measurable null union. Redefining all $f_n$ and $f$ to be zero on that union reduces the proof to pointwise monotonicity and convergence without changing any integral.

Set

$$\alpha = \sup_n\int_Xf_n\,\mathrm d\mu.$$

Since $f_n\leq f$, monotonicity gives

$$\alpha\leq\int_Xf\,\mathrm d\mu.$$

For the reverse inequality, choose any nonnegative simple function $s\leq f$ and any number $c$ with $0<c<1$. Define

$$E_n=\{x\in X:f_n(x)\geq c\,s(x)\}.$$

These sets are measurable and increase with $n$. Moreover, they exhaust $X$: if $s(x)=0$, then $x\in E_n$ for every $n$; if $s(x)>0$, then $f_n(x)\uparrow f(x)\geq s(x)>c\,s(x)$, so $x$ eventually belongs to $E_n$. On $E_n$,

$$f_n\geq c\,s,$$

and hence

$$\int_Xf_n\,\mathrm d\mu \geq c\int_{E_n}s\,\mathrm d\mu.$$

If $s=\sum_j a_j\mathbf1_{A_j}$ in disjoint-level-set form, continuity from below of the measure gives

$$\int_{E_n}s\,\mathrm d\mu = \sum_j a_j\mu(A_j\cap E_n) \longrightarrow \sum_j a_j\mu(A_j) = \int_Xs\,\mathrm d\mu.$$

Therefore

$$\alpha\geq c\int_Xs\,\mathrm d\mu.$$

Letting $c\uparrow1$ and then taking the supremum over every nonnegative simple $s\leq f$ yields

$$\alpha\geq\int_Xf\,\mathrm d\mu.$$

Together with the first inequality, this proves the theorem.

Consequences

The theorem recovers continuity from below by taking $f_n=\mathbf1_{A_n}$ for $A_n\uparrow A$. It also gives a useful decreasing version, but only with a finiteness hypothesis:

$$0\leq f_n\downarrow f, \qquad \int_Xf_1\,\mathrm d\mu<\infty \quad\Longrightarrow\quad \int_Xf_n\,\mathrm d\mu\downarrow\int_Xf\,\mathrm d\mu.$$

Indeed, $f_1-f_n\uparrow f_1-f$, so monotone convergence applies before one subtracts the finite number $\int f_1\,\mathrm d\mu$. More precisely, integrability makes $\{f_1=\infty\}$ null; redefine all the functions to be zero there before forming these differences. The finiteness condition cannot simply be dropped: on $\mathbb R$ with Lebesgue measure,

$$f_n=\mathbf1_{[n,\infty)} \downarrow0$$

pointwise, while $\int_{\mathbb R}f_n\,\mathrm dx=\infty$ for every $n$.

For nonnegative measurable functions $h_j$, let $S_n=\sum_{j=1}^nh_j$. Since $S_n\uparrow\sum_{j=1}^{\infty}h_j$,

$$\int_X\sum_{j=1}^{\infty}h_j\,\mathrm d\mu = \sum_{j=1}^{\infty}\int_Xh_j\,\mathrm d\mu,$$

possibly with both sides equal to $+\infty$. This is a series theorem on one measure space. The Tonelli theorem for a product measure belongs to the next page.

Why monotonicity matters

On $(0,1)$ with Lebesgue measure, define

$$u_n(x)=n\mathbf1_{(0,1/n)}(x).$$

Then $u_n\to0$ pointwise, but

$$\int_0^1u_n(x)\,\mathrm dx=1$$

for every $n$. The mass concentrates into a shrinking interval rather than disappearing. The sequence is nonnegative but not increasing, so monotone convergence does not apply.

This example also has no common integrable dominator. If $g\geq u_n$ almost everywhere for every $n$, then, outside one common null set,

$$g(x)\geq k \qquad \text{for } x\in\left(\frac{1}{k+1},\frac{1}{k}\right).$$

Consequently,

$$\int_0^1g(x)\,\mathrm dx \geq \sum_{k=1}^{\infty} k\left(\frac1k-\frac1{k+1}\right) = \sum_{k=1}^{\infty}\frac1{k+1} =\infty.$$

Thus dominated convergence cannot repair the missing monotonicity here.

Fatou’s lemma

The lower limit of a sequence is

$$\liminf_{n\to\infty}f_n(x) = \lim_{n\to\infty}\inf_{k\geq n}f_k(x).$$

It records the eventual lower envelope rather than requiring the sequence to converge.

Fatou’s lemma. If $f_n:X\to[0,\infty]$ is measurable for every $n$, then

$$\int_X\liminf_{n\to\infty}f_n\,\mathrm d\mu \leq \liminf_{n\to\infty}\int_Xf_n\,\mathrm d\mu.$$

See Lin 2021, Lecture 11, p. 57, PDF and Tao 2011, Corollary 1.4.47, p. 110, author-hosted preliminary PDF for the standard theorem and proof.

Proof

Define

$$g_n=\inf_{k\geq n}f_k.$$

A countable infimum of measurable functions is measurable, and

$$0\leq g_n\uparrow\liminf_{k\to\infty}f_k.$$

Monotone convergence therefore gives

$$\int_X\liminf_{k\to\infty}f_k\,\mathrm d\mu = \lim_{n\to\infty}\int_Xg_n\,\mathrm d\mu.$$

For every $k\geq n$, the inequality $g_n\leq f_k$ implies

$$\int_Xg_n\,\mathrm d\mu \leq \inf_{k\geq n}\int_Xf_k\,\mathrm d\mu.$$

Taking $n\to\infty$ proves the claim.

Fatou gives an inequality, not usually an equality. On $\mathbb R$,

$$b_n=\mathbf1_{[n,n+1]}$$

converges pointwise to zero, while every integral is one. Hence

$$0 = \int_{\mathbb R}\liminf_nb_n\,\mathrm dx < \liminf_n\int_{\mathbb R}b_n\,\mathrm dx =1.$$

Nonnegativity is essential. For

$$v_n(x)=-n\mathbf1_{(0,1/n)}(x) \qquad\text{on }(0,1),$$

the pointwise limit is zero but every integral equals $-1$. Applying the displayed Fatou inequality unchanged would assert the false statement $0\leq-1$.

There is a valid signed extension. Let the extended-real functions $f_n$ be measurable, and suppose one integrable function $h$ satisfies $f_n\geq h$ almost everywhere for every $n$. Redefine the functions on the common exceptional null set so that the inequalities hold pointwise, and apply Fatou to the nonnegative functions $f_n-h$. Because $\int h\,\mathrm d\mu$ is finite, all the displayed extended integrals are well defined and subtraction is legal:

$$\int_X\liminf_nf_n\,\mathrm d\mu \leq \liminf_n\int_Xf_n\,\mathrm d\mu.$$

The same lower bound $h$ must work for the whole sequence.

Dominated convergence

Monotonicity is not required when a single integrable function controls the absolute values.

Dominated convergence theorem. Let $f_n:X\to\mathbb C$ and $f:X\to\mathbb C$ be measurable. Suppose

$$f_n\longrightarrow f \quad\text{almost everywhere}$$

and there is one measurable function $g:X\to[0,\infty]$ such that

$$|f_n|\leq g \quad\text{almost everywhere for every }n, \qquad \int_Xg\,\mathrm d\mu<\infty.$$

Then $f\in L^1(\mu)$ and

$$\int_X|f_n-f|\,\mathrm d\mu\longrightarrow0.$$

In particular,

$$\int_Xf_n\,\mathrm d\mu \longrightarrow \int_Xf\,\mathrm d\mu.$$

See Lin 2021, Lecture 12, pp. 61–64, PDF and Tao 2011, Theorem 1.4.49, pp. 111–112, author-hosted preliminary PDF for independent proof treatments.

The real-valued theorem is the special case $f_n,f:X\to\mathbb R$. Measurability of the stated limit $f$ is explicit: on an incomplete measure space, arbitrary values assigned on a subset of a null set need not define a measurable function.

Proof

Take the union of the null sets on which convergence or one of the bounds fails, and redefine $f_n$ and $f$ to be zero there. These measurable redefinitions preserve all integrals and make convergence and domination hold pointwise. Consequently,

$$|f| = \lim_n|f_n| \leq g.$$

Thus $f$ and every $f_n$ are integrable, and

$$0\leq|f_n-f|\leq2g.$$

Apply Fatou’s lemma to the nonnegative functions

$$2g-|f_n-f|.$$

Because $|f_n-f|\to0$ almost everywhere and $\int g<\infty$,

$$\begin{aligned} 2\int_Xg\,\mathrm d\mu &= \int_X\liminf_n\bigl(2g-|f_n-f|\bigr)\,\mathrm d\mu \\ &\leq \liminf_n \left( 2\int_Xg\,\mathrm d\mu - \int_X|f_n-f|\,\mathrm d\mu \right) \\ &= 2\int_Xg\,\mathrm d\mu - \limsup_n\int_X|f_n-f|\,\mathrm d\mu. \end{aligned}$$

It follows that

$$\limsup_n\int_X|f_n-f|\,\mathrm d\mu\leq0.$$

The integrals are nonnegative, so they converge to zero. Finally,

$$\left| \int_Xf_n\,\mathrm d\mu-\int_Xf\,\mathrm d\mu \right| \leq \int_X|f_n-f|\,\mathrm d\mu \longrightarrow0.$$

This completes the proof.

What “dominated” means

The function $g$ is not merely a pointwise bound selected after $n$ is fixed. It must be

• the same function for every $n$;
• independent of any cutoff parameter being removed;
• measurable and nonnegative;
• integrable with respect to the same fixed measure $\mu$.

For the concentrating sequence $u_n=n\mathbf1_{(0,1/n)}$, the natural envelopes depend on $n$, and no common integrable envelope exists.

Bounded convergence. If the functions $f_n,f:X\to\mathbb C$ are measurable, $\mu(X)<\infty$, $f_n\to f$ almost everywhere, and $|f_n|\leq M$ for one finite constant $M$, then dominated convergence applies with

$$g=M\mathbf1_X, \qquad \int_Xg\,\mathrm d\mu=M\mu(X)<\infty.$$

The finite-measure hypothesis cannot be omitted. The wandering functions $\mathbf1_{[n,n+1]}$ on $\mathbb R$ are uniformly bounded by one and converge pointwise to zero, but their integrals remain one.

Absolutely integrable series

For measurable real- or complex-valued functions $h_j$, suppose

$$\sum_{j=1}^{\infty} \int_X|h_j|\,\mathrm d\mu < \infty.$$

Monotone convergence applied to $g_n=\sum_{j=1}^n|h_j|$ gives

$$g=\sum_{j=1}^{\infty}|h_j|\in L^1(\mu).$$

In particular, the original series converges absolutely almost everywhere. Its partial sums $H_n=\sum_{j=1}^nh_j$ satisfy $|H_n|\leq g$, so dominated convergence yields

$$\int_X\sum_{j=1}^{\infty}h_j\,\mathrm d\mu = \sum_{j=1}^{\infty}\int_Xh_j\,\mathrm d\mu.$$

Absolute integrability is what prevents an illegal rearrangement of $\infty-\infty$.

Which theorem applies?

Each theorem answers a different question.

Available information	Result	Conclusion
$0\leq f_n\uparrow f$ a.e.	Monotone convergence	Equality of limit and integral; $+\infty$ allowed
$f_n\geq0$ only	Fatou	One-sided lower-limit inequality
$f_n\to f$ a.e. and $\lvert f_n\rvert\leq g\in L^1$	Dominated convergence	$L^1$ convergence and convergence of integrals
$f_n\to f$ a.e., $\lvert f_n\rvert\leq M$, and $\mu(X)<\infty$	Bounded convergence	Dominated convergence with $g=M\mathbf1_X$
Pointwise or a.e. convergence alone	No general theorem	A separate estimate or additional hypothesis is needed

These hypotheses are sufficient, not necessary. Failure to find a dominating function does not prove that the integrals fail to converge; it only means that this particular proof is unavailable. Conversely, naming a theorem is not enough: every hypothesis must concern the actual sequence, measure, and domain in the proposed limit.

Convergence in measure

A sequence of measurable functions converges in measure to $f$ if, for every $\varepsilon>0$,

$$\mu\!\left( \{x\in X:|f_n(x)-f(x)|>\varepsilon\} \right) \longrightarrow0.$$

This mode of convergence ignores errors on sets whose measure tends to zero. It is related to several notions already encountered, but every implication below has its own hypotheses. Definitions, examples, and the subsequence criterion appear in Tao 2011, §1.5, pp. 115–125, author-hosted preliminary PDF.

From L¹ convergence to convergence in measure

On the exceptional set

$$E_{n,\varepsilon} = \{|f_n-f|>\varepsilon\},$$

one has

$$\varepsilon\mathbf1_{E_{n,\varepsilon}} \leq |f_n-f|.$$

Integration gives the estimate

$$\mu(E_{n,\varepsilon}) \leq \frac1\varepsilon \int_X|f_n-f|\,\mathrm d\mu.$$

Therefore $L^1$ convergence always implies convergence in measure, whether or not $\mu(X)$ is finite.

From almost-everywhere convergence

If $f_n\to f$ almost everywhere, then for each fixed $\varepsilon>0$,

$$\mathbf1_{E_{n,\varepsilon}}\longrightarrow0 \quad\text{almost everywhere}.$$

When $\mu(X)<\infty$, the integrable function $\mathbf1_X$ dominates these indicators. Dominated convergence then gives

$$\mu(E_{n,\varepsilon}) = \int_X\mathbf1_{E_{n,\varepsilon}}\,\mathrm d\mu \longrightarrow0.$$

Finite total measure is essential. On $\mathbb R$, $\mathbf1_{[n,n+1]}\to0$ pointwise, but for $0<\varepsilon<1$ every exceptional set has measure one.

Neither convergence in measure nor pointwise convergence alone implies $L^1$ convergence. The concentrating sequence

$$u_n=n\mathbf1_{(0,1/n)}$$

converges in measure to zero because its support has measure $1/n$, but $\int|u_n|=1$ for every $n$.

There is also no converse from convergence in measure to pointwise convergence of the full sequence. On $[0,1)$, for $k\geq0$ and $0\leq j<2^k$, define

$$t_{2^k+j} = \mathbf1_{[j/2^k,(j+1)/2^k)}.$$

The support measure in the $k$th block is $2^{-k}$, so $t_n\to0$ in measure. Every point belongs to exactly one interval in each block, however, and therefore sees infinitely many values equal to one and infinitely many equal to zero. The sequence has no pointwise limit.

A useful partial converse survives: every sequence converging in measure has a subsequence converging almost everywhere to the same limit. Choose $n_j$ so that

$$\mu\!\left( \{|f_{n_j}-f|>2^{-j}\} \right) < 2^{-j}.$$

The union of these exceptional sets over $j\geq m$ has measure at most $2^{1-m}$. Their limsup is therefore null, and outside it $|f_{n_j}-f|\leq2^{-j}$ eventually. Full $L^p$, norm, and weak convergence are developed on the later $L^p$ page.

The Radon–Nikodym theorem

Integration can also describe one measure relative to another. For positive measures $\mu$ and $\nu$ on the same measurable space, write

$$\nu\ll\mu$$

and say that $\nu$ is absolutely continuous with respect to $\mu$ when

$$\mu(A)=0\quad\Longrightarrow\quad\nu(A)=0 \qquad(A\in\Sigma).$$

Radon–Nikodym theorem. If $\mu$ and $\nu$ are positive sigma-finite measures and $\nu\ll\mu$, then there is a measurable function $h:X\to[0,\infty]$, unique $\mu$-almost everywhere, such that

$$\nu(A)=\int_Ah\,\mathrm d\mu \qquad \text{for every }A\in\Sigma.$$

The function is denoted

$$h=\frac{\mathrm d\nu}{\mathrm d\mu}$$

and is called the Radon–Nikodym derivative. This page uses the theorem as a stated result. Its existence proof uses a different set of tools: standard proofs construct densities on finite-measure pieces by signed-measure or functional-representation arguments and patch them consistently over a sigma-finite exhaustion. See Folland 1999, Theorem 3.8, pp. 90–92 and Tao 2009, Theorem 2 and Corollary 1 for complete existence proofs.

The almost-everywhere uniqueness is short once existence is known. If $h$ and $k$ are two densities, choose a measurable cover $X_m\uparrow X$ with $\nu(X_m)<\infty$. Testing the equality of the two induced measures on

$$\{h>k\}\cap X_m$$

shows that the integral of the positive difference $h-k$ there is zero. Thus $h\leq k$ almost everywhere on every $X_m$; interchanging $h$ and $k$ gives equality almost everywhere.

Absolute continuity is indispensable. The Dirac measure $\delta_0$ has no density with respect to Lebesgue measure because $\{0\}$ is Lebesgue-null but has $\delta_0$-measure one. Sigma-finiteness is also a theorem hypothesis, not a decorative assumption.

For a finite signed measure absolutely continuous with respect to a sigma-finite positive measure, applying the positive theorem to the Jordan decomposition produces an integrable real density. That version supports the construction of conditional expectation on Probability Spaces, Random Variables, and Conditional Expectation; see Sheffield 2014, Lecture 25, slide 6, PDF.

A finite Gaussian amplitude cutoff

The convergence theorems can rigorously remove an amplitude cutoff from the finite-dimensional Gaussian introduced on the preceding page. Fix $N<\infty$, let $K$ be a real symmetric positive-definite $N\times N$ matrix, and let $P:\mathbb R^N\to\mathbb C$ be a polynomial. For

$$B_R=\{q\in\mathbb R^N:\|q\|\leq R\},$$

define

$$I_R(P) = \int_{\mathbb R^N} \mathbf1_{B_R}(q)\, P(q) e^{-\frac12q^{\mathsf T}Kq} \,\mathrm d^Nq.$$

The finite-dimensional Gaussian and polynomial-observable setting is standard; see Zinn-Justin 2021, Chapter 1, §§1.1–1.2, pp. 1–3. The cutoff estimates below are derived directly.

The integration setting is ordinary Lebesgue measure on $\mathbb R^N$, and the Euclidean weight $e^{-S_E}$ is positive when $P=1$. Let

$$\kappa_{\min}>0$$

be the smallest eigenvalue of $K$.

The partition function: use monotone convergence

For $P=1$, the cutoff integrands are nonnegative and increase as $R$ increases. Taking integer radii gives

$$\mathbf1_{B_n}(q)e^{-\frac12q^{\mathsf T}Kq} \uparrow e^{-\frac12q^{\mathsf T}Kq}.$$

Monotone convergence therefore yields

$$I_n(1)\uparrow Z_K, \qquad Z_K = \int_{\mathbb R^N} e^{-\frac12q^{\mathsf T}Kq} \,\mathrm d^Nq = \frac{(2\pi)^{N/2}}{\sqrt{\det K}}.$$

Because $I_R(1)$ is monotone in $R$ and integer balls exhaust $\mathbb R^N$, the same limit holds as the continuous parameter $R\to\infty$.

Polynomial observables: use dominated convergence

A polynomial has a global growth bound

$$|P(q)| \leq C(1+\|q\|^m)$$

for some constants $C>0$ and integer $m\geq0$. Positive definiteness gives

$$q^{\mathsf T}Kq \geq \kappa_{\min}\|q\|^2.$$

Since

$$(1+r^m)e^{-\kappa_{\min}r^2/4}$$

is bounded for $r\geq0$, there is a constant $C'$ such that

$$\begin{aligned} |P(q)|e^{-\frac12q^{\mathsf T}Kq} &\leq C(1+\|q\|^m) e^{-\kappa_{\min}\|q\|^2/2} \\ &\leq C'e^{-\kappa_{\min}\|q\|^2/4}. \end{aligned}$$

The last function is integrable on $\mathbb R^N$ by the finite-dimensional Gaussian formula. It is independent of $R$. For every sequence $R_n\to\infty$,

$$\mathbf1_{B_{R_n}}(q) P(q)e^{-\frac12q^{\mathsf T}Kq} \longrightarrow P(q)e^{-\frac12q^{\mathsf T}Kq}$$

pointwise, so dominated convergence gives

$$I_{R_n}(P) \longrightarrow I(P) \equiv \int_{\mathbb R^N} P(q)e^{-\frac12q^{\mathsf T}Kq} \,\mathrm d^Nq.$$

Every sequence $R_n\to\infty$ has the same limit; the sequential criterion for a real parameter therefore gives $I_R(P)\to I(P)$ as $R\to\infty$.

Since $Z_K>0$ and $I_R(1)>0$ for $R>0$,

$$\frac{I_R(P)}{I_R(1)} \longrightarrow \frac{I(P)}{Z_K}.$$

For the check $P(q)=q_aq_b$, the finite Gaussian covariance calculation on Measures and Measurable Functions gives

$$\frac{I(q_aq_b)}{Z_K} =(K^{-1})_{ab}.$$

This is a controlled limit, but its scope is narrow:

• $N$ remains fixed and finite.
• The removed restriction is a field-amplitude cutoff, not a mode or ultraviolet regulator.
• The argument uses a positive Euclidean Gaussian and $K>0$.
• Sending $N\to\infty$ changes the underlying spaces; the theorem above cannot be applied without constructing a common measurable setting and uniform bounds.
• No continuum, interacting, Lorentzian, or renormalized conclusion follows.

The physical finite-mode construction is developed on Regulated Bosonic Field Integrals.

Common pitfalls

Almost-everywhere convergence is not enough. The spike $n\mathbf1_{(0,1/n)}$ converges almost everywhere to zero while retaining unit integral. Check monotonicity, nonnegativity, and domination separately.

An $n$-dependent bound is not a dominator. Dominated convergence needs one integrable $g$ that bounds every member of the sequence with respect to the same measure.

A constant bound need not be integrable. On an infinite-measure space, $|f_n|\leq1$ does not supply $1\in L^1(\mu)$. Bounded convergence requires $\mu(X)<\infty$.

Monotone convergence may return infinity. The theorem does not assert that the limiting function is integrable; it asserts equality in $[0,\infty]$.

Fatou cannot be applied unchanged to signed functions. Nonnegativity, or one common integrable lower bound, is required.

Extended integrals do not permit $\infty-\infty$. Establish absolute integrability before using unrestricted linearity or subtracting two infinite contributions.

A changing cutoff can change the measure space. Removing an amplitude cutoff inside a fixed $\mathbb R^N$ is different from sending $N\to\infty$. The latter is not an immediate application of dominated convergence.

A failed theorem test is not a divergence proof. The convergence theorems provide sufficient conditions. A different estimate may still establish the desired limit.

Where the next pages begin

• Product Measures, Fubini–Tonelli, and Change of Variables develops product sigma-algebras, iterated integration, and Jacobians.
• $L^p$ Spaces, Inequalities, and Weak Convergence develops normed equivalence classes, Hölder and Minkowski inequalities, density, and weak convergence.
• Probability Spaces, Random Variables, and Conditional Expectation applies integration and the Radon–Nikodym theorem to expectation and conditioning.
• Regulated Bosonic Field Integrals gives the developed physical finite-mode application.
• Foundations develops the finite-regulator setting in which these integration tools first enter the physical theory.

Exercises

A simple integral. Let $A$ and $B$ be disjoint measurable sets with $\mu(A)=2$ and $\mu(B)=4$. Compute the integral of

$$s=3\mathbf1_A+\frac12\mathbf1_B.$$

Solution

The sets are already disjoint level sets, so the definition gives

$$\int_Xs\,\mathrm d\mu = 3\mu(A)+\frac12\mu(B) = 3\cdot2+\frac12\cdot4 =8.$$

Diagnose the spike. For $u_n=n\mathbf1_{(0,1/n)}$ on $(0,1)$, determine whether monotone convergence, Fatou’s lemma, or dominated convergence proves convergence of the integrals to the integral of the pointwise limit.

Solution

The pointwise limit is zero and all $u_n$ are nonnegative, but the sequence is not increasing, so monotone convergence does not apply. Fatou applies and gives only the true lower bound

$$0 \leq \liminf_n\int_0^1u_n\,\mathrm dx =1.$$

Dominated convergence does not apply because there is no common integrable dominator. Indeed, domination on $(1/(k+1),1/k)$ forces $g\geq k$ almost everywhere there, and the resulting harmonic lower bound makes $\int g=\infty$. The integrals therefore remain one rather than converging to zero.

Recover Fatou from monotone convergence. Given nonnegative measurable $f_n$, identify an increasing sequence to which monotone convergence applies, and derive Fatou’s inequality.

Solution

Take

$$g_n=\inf_{k\geq n}f_k.$$

Then $g_n\uparrow\liminf_kf_k$. Monotone convergence and the inequalities $g_n\leq f_k$ for every $k\geq n$ give

$$\begin{aligned} \int_X\liminf_kf_k\,\mathrm d\mu &= \lim_n\int_Xg_n\,\mathrm d\mu \\ &\leq \lim_n\inf_{k\geq n} \int_Xf_k\,\mathrm d\mu \\ &= \liminf_k\int_Xf_k\,\mathrm d\mu. \end{aligned}$$

Choose the Gaussian theorem. In the finite Gaussian cutoff example, which theorem removes the cutoff for $P=1$, and which theorem handles a general polynomial $P$? State the decisive hypothesis in each case.

Solution

For $P=1$, the nonnegative functions

$$\mathbf1_{B_n}(q)e^{-q^{\mathsf T}Kq/2}$$

increase pointwise to the full Gaussian weight, so monotone convergence is the direct theorem. For a polynomial $P$, the integrands need not be nonnegative or monotone. Dominated convergence applies because

$$\left| \mathbf1_{B_R}(q)P(q)e^{-q^{\mathsf T}Kq/2} \right| \leq C'e^{-\kappa_{\min}\|q\|^2/4},$$

and the right side is one $R$-independent integrable Gaussian.

References

• Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Wiley (1999), Folland 1999, §§2.2–2.4 and 3.2, pp. 49–63 and 88–92, publisher record. Integration, convergence, and the Lebesgue–Radon–Nikodym theorem.
• Andrew Lin, lecture notes for Casey Rodriguez’s 18.102 Introduction to Functional Analysis, Spring 2021, MIT OpenCourseWare: Lin 2021, Lecture 10, “Simple Functions,” pp. 47–52, PDF; Lin 2021, Lecture 11, “The Lebesgue Integral of a Nonnegative Function and Convergence Theorems,” pp. 53–58, PDF; and Lin 2021, Lecture 12, “Lebesgue Integrable Functions, the Lebesgue Integral and the Dominated Convergence Theorem,” pp. 59–64, PDF. Simple and complex integration, monotone convergence, Fatou’s lemma, and dominated convergence.
• Scott Sheffield, 18.175 Theory of Probability, Lecture 25, MIT OpenCourseWare (2014), Sheffield 2014, Lecture 25, slide 6, PDF. Probability-facing Radon–Nikodym statement and conditional expectation.
• Terence Tao, 245B, Notes 1: Signed Measures and the Radon–Nikodym–Lebesgue Theorem (2009), Tao 2009, Theorem 2 and Corollary 1, author-hosted notes. Open proof reference for the Radon–Nikodym theorem and its signed-measure extension.
• Terence Tao, An Introduction to Measure Theory, Graduate Studies in Mathematics 126, American Mathematical Society (2011), Tao 2011, §§1.3–1.5, pp. 50–72 and 105–125, author-hosted preliminary PDF. Integration, convergence theorems, convergence in measure, and counterexamples.
• Jean Zinn-Justin, Quantum Field Theory and Critical Phenomena, 5th ed., Oxford University Press (2021), Zinn-Justin 2021, Chapter 1, §§1.1–1.2, pp. 1–3, OUP. Finite-dimensional Gaussian integrals, sources, and polynomial expectations.