Exceptional Fibers with Infinite Section Integrals

Infinite Section Integrals Do Not Contradict Finite Total Integrals

For a function on a product space, fixing one coordinate gives a section, and integrating that section produces a function of the remaining coordinate. That section integral may equal $+\infty$ at some parameters even when the original function has finite integral. The key distinction is between a statement about every parameter and one that holds almost everywhere: a null set of exceptional parameters can carry infinite section integrals without changing the outer integral.

Throughout, let $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ be sigma-finite measure spaces. Their product is equipped with $\mathcal{M}\otimes\mathcal{N}$ and $\mu\times\nu$. The functions considered below are measurable and take values in $[0,\infty]$. For such a function $f$, its section at $x\in X$ is $f_x(y)=f(x,y)$, and its section integral is the extended-real-valued quantity $G(x)=\int_Y f(x,y)\,d\nu(y)$.

Tonelli’s theorem (the form used here): If $f:X\times Y\to[0,\infty]$ is $\mathcal{M}\otimes\mathcal{N}$-measurable and both measure spaces are sigma-finite, then $G(x)=\int_Y f(x,y)\,d\nu(y)$ is measurable, and $$ \int_{X\times Y} f\,d(\mu\times\nu) =\int_X G(x)\,d\mu(x) =\int_X\int_Y f(x,y)\,d\nu(y)\,d\mu(x). $$ The equalities are in $[0,\infty]$. The corresponding statement holds with the roles of $X$ and $Y$ reversed.

Tonelli does not say that $G(x)$ is finite for every $x$. It says that $G$ is measurable and that its integral gives the product integral. If that integral is finite, then the infinite values of $G$ can occur only on a $\mu$-null set, as the next result makes precise.

Finite Total Integral Forces Finite Sections Almost Everywhere

Proposition: Under the sigma-finiteness and measurability hypotheses of Tonelli’s theorem, suppose $f\ge0$ and $\int_{X\times Y}f\,d(\mu\times\nu)<\infty$. Then $\int_Y f(x,y)\,d\nu(y)<\infty$ for $\mu$-almost every $x$. The claim allows exceptional parameters; it does not assert finiteness at every $x$.

Proof. By Tonelli’s theorem, $G(x)=\int_Y f(x,y)\,d\nu(y)$ is measurable and $\int_XG\,d\mu=I$, where $I=\int_{X\times Y}f\,d(\mu\times\nu)<\infty$. For each positive integer $n$, let $E_n=\{x:G(x)>n\}$. Since $G\ge n\chi_{E_n}$, monotonicity of the nonnegative integral gives $n\mu(E_n)\le I$, so $\mu(E_n)\le I/n$. Also, $\{G=\infty\}\subseteq E_n$ for every $n$. Thus $0\le\mu(\{G=\infty\})\le I/n$ for every $n$, and letting $n\to\infty$ gives $\mu(\{G=\infty\})=0$. Therefore $G$ is finite almost everywhere. $\square$

The estimate also covers $I=0$: in that case each $E_n$ has measure zero. A finite integral therefore permits exceptional infinite sections but rules out a positive-measure set of them.

Worked Example: One Infinite Section and a Finite Product Integral

Let $X=[-1,1]$ with Lebesgue measure and $Y=[0,\infty)$ with Lebesgue measure. Define $f:X\times Y\to[0,\infty]$ by setting $f(0,y)=1$ for every $y\ge0$, and, when $x\ne0$, setting $f(x,y)=1$ if $0\le y<x^2$ and $f(x,y)=0$ otherwise. The defining sets are Borel, so $f$ is product-measurable. Both spaces are sigma-finite.

First fix $x$. If $x=0$, the section is constantly $1$ on the infinite interval $[0,\infty)$, so $G(0)=\int_0^\infty1\,dy=\infty$. If $x\ne0$, the section is $1$ on $[0,x^2)$, whose length is $x^2$, and zero elsewhere. Hence $G(x)=x^2$ for $x\ne0$. Integrating the section function gives

\int_{-1}^1G(x)\,dx =\int_{-1}^0x^2\,dx+\int_0^1x^2\,dx =\frac13+\frac13 =\frac23.

The value $G(0)=\infty$ does not alter this integral because $\{0\}$ has measure zero. To check the result in the other order, fix $y$. At $y=0$, the condition $0\le y<x^2$ holds for every $x\ne0$, giving an $x$-section integral of $2$. For $0<y<1$, it holds exactly when $|x|>\sqrt y$, a subset of $[-1,1]$ of length $2(1-\sqrt y)$. For $y\ge1$, it holds for no $x$. The value at $x=0$ contributes only on a singleton in each $x$-integral. Therefore the reversed integral is

\int_0^1 2(1-\sqrt y)\,dy =2\left(1-\frac23\right) =\frac23.

Check: both orders give $2/3$, as Tonelli requires. The section integral is infinite at exactly the exceptional parameter $x=0$, and finite at every $x\ne0$.

Changing Values on a Null Fiber

The example modifies the function along the fiber $\{0\}\times Y$. Under sigma-finiteness of $Y$, that whole fiber has product measure zero. More generally, if $N\in\mathcal{M}$ and $\mu(N)=0$, then $N\times Y$ is product-null: cover $Y$ by measurable sets $Y_j$ of finite $\nu$-measure, observe that $(\mu\times\nu)(N\times Y_j)=\mu(N)\nu(Y_j)=0$, and take the countable union.

Proposition: If two nonnegative $\mathcal{M}\otimes\mathcal{N}$-measurable functions agree outside a product-null set, then their product integrals are equal, with $+\infty$ allowed.

Proof. Let the functions be $f$ and $g$, and let $Z$ be a product-null measurable set outside which they agree. The nonnegative functions $f\chi_Z$ and $g\chi_Z$ each have integral zero: every nonnegative simple function below either one is supported on $Z$, so each of its level sets has measure zero and its simple-function integral is zero. On $Z^c$, the functions agree. Decomposing each integral over $Z$ and $Z^c$ therefore gives $\int f\,d(\mu\times\nu)=\int g\,d(\mu\times\nu)$. $\square$

Worked Example: Infinite Section Integrals on a Countable Null Set

Take $X=[0,1]$, $Y=[0,\infty)$, both with Lebesgue measure, and let $N=\mathbb{Q}\cap[0,1]$. Define $f(x,y)=\chi_{N\times[0,\infty)}(x,y)$. Since $N$ is countable, it is measurable and null; the displayed rectangle is product-measurable.

If $x\in N$, then $f(x,y)=1$ for all $y\ge0$, so $\int_Y f(x,y)\,dy=\infty$. If $x\notin N$, the section is zero, and its integral is $0$. Consequently the section integral function is infinite on $N$ and zero on $X\setminus N$. Its integral over $X$ is $0$, since $N$ is null. Reversing the order also gives zero: for every fixed $y\ge0$, the $x$-section is $\chi_N$, whose integral is $\mu(N)=0$, so the integral over $Y$ of these section integrals is $\int_0^\infty0\,dy=0$.

Check: Tonelli gives product integral $0$ in both orders. The exceptional set is now countable and dense in $[0,1]$, but it still has measure zero; density does not change the calculation.

Unbounded Finite Sections Can Also Occur

Finiteness almost everywhere does not imply that section integrals have a uniform finite bound. They may grow without bound near an exceptional parameter while remaining integrable. The following example combines that behavior with one genuinely infinite section.

Worked Example: An Unbounded Section Integral with Finite Total

Let $X=[0,1]$, $Y=[0,\infty)$, with Lebesgue measure. Set $f(0,y)=1$ for every $y\ge0$. For $x>0$, set $f(x,y)=x^{-1/2}$ when $0\le y<1$, and set it to zero when $y\ge1$. This is a nonnegative product-measurable function on sigma-finite spaces.

At $x=0$, the section integral is $\int_0^\infty1\,dy=\infty$. At each $x>0$, it is $x^{-1/2}$, because the nonzero section has $y$-length $1$. Thus the section integral is finite almost everywhere but unbounded as $x$ approaches $0$. Its integral is computed by increasing the intervals $[1/n,1]$ to $(0,1]$; the Monotone Convergence Theorem applies to the corresponding nonnegative truncations:

\int_0^1G(x)\,dx =\lim_{n\to\infty}\int_{1/n}^1x^{-1/2}\,dx =\lim_{n\to\infty}2\left(1-\frac1{\sqrt n}\right) =2.

For the reverse order, at every $0\le y<1$, the $x$-section equals $x^{-1/2}$ for $x>0$, while its value at the single point $x=0$ does not affect its integral. That integral is $2$. For $y\ge1$, the section is zero except possibly at $x=0$, again a null set, so its integral is zero. The reversed integral is therefore $\int_0^1 2\,dy=2$.

Check: the two orders both give $2$. The section integral diverges at $x=0$, grows arbitrarily large for positive $x$ near zero, and nevertheless has finite integral.

What Tonelli and Fubini Do—and Do Not—Assert

In the first and third examples, the product integral is finite, so the nonnegative function is integrable on the product space. Fubini’s theorem is therefore consistent with the infinite section at one parameter: its section conclusions are almost-everywhere conclusions. Neither Fubini nor Tonelli asserts that every individual section is integrable. Sigma-finiteness is part of the product-integration hypotheses used here, and the integral identity alone should not be read as a pointwise statement about all parameters.

A common mistake is to treat an infinite value at one parameter as if it made the integral of the section-integral function infinite. In the first example, the incorrect calculation would be: “$G(0)=\infty$, so $\int_{-1}^1G(x)\,dx=\infty$.” This confuses a value at a point with a contribution over a set of positive measure. The correction is to use the actual measurable section-integral function: it equals $x^2$ away from $0$, and its value at $0$ does not affect its Lebesgue integral. In particular, the function equal to $+\infty$ at $0$ and zero elsewhere has integral zero, not infinity.

Conversely, merely knowing that section integrals are finite almost everywhere does not guarantee a finite total integral: on $X=\mathbb{R}$, $Y=[0,1]$, the function $f(x,y)=1$ has section integral $1$ for every $x$, but its product integral is infinite because $X$ has infinite measure. The hypotheses and conclusion must be kept distinct: finite total integral implies finite section integrals almost everywhere, but almost-everywhere finiteness by itself does not imply a finite total.

Check Your Understanding

Use Tonelli’s theorem with its measurability and sigma-finiteness hypotheses stated whenever you apply it.

In the first worked example, what is the section integral at $x=0$, and why does it not make the total integral infinite?
For the function in the countable-null-set example, identify the set of parameters with infinite section integral and compute its measure.
Why does finite product integral imply that the set of parameters with infinite section integral is null?
In the third worked example, explain why the section integral is unbounded but has integral $2$.
Give an example in which section integrals are finite everywhere but the product integral is infinite.