Parameter Integrals and Exchange of Integrals

Parameter Integrals as Sections

A parameter integral is obtained by fixing one variable in a function of several variables and integrating over another. For example, a function $F(s)=\int f(s,t)\,dt$ is defined by integrating the $t$-section while $s$ is held fixed. Tonelli’s and Fubini’s theorems give conditions under which these section integrals are measurable and the order of integration can be exchanged. The distinction matters: nonnegative functions may have infinite integrals, while exchanging the order for signed or complex functions requires absolute integrability.

Let $(X,\mathcal M,\mu)$ and $(Y,\mathcal N,\nu)$ be sigma-finite measure spaces; neither is assumed complete or finite. We use their product measure $\mu\times\nu$ on $\mathcal M\otimes\mathcal N$. For a measurable function $f:X\times Y\to[0,\infty]$, its section at $y\in Y$ is $f_y(x)=f(x,y)$. The parameter integral is the extended-valued function $G(y)=\int_X f(x,y)\,d\mu(x)$. Tonelli’s theorem says that $G$ is measurable and that its integral equals the product integral, with $+\infty$ allowed. For a real- or complex-valued measurable function, Fubini’s theorem instead requires $\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty$.

Parameter-integral rule: If $f:X\times Y\to[0,\infty]$ is $\mathcal M\otimes\mathcal N$-measurable, Tonelli’s theorem gives a measurable $G(y)=\int_X f(x,y)\,d\mu(x)$ and $$ \int_Y G(y)\,d\nu(y)=\int_{X\times Y}f\,d(\mu\times\nu)=\int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x). $$ The integrals may be infinite. If $f$ is real- or complex-valued and $\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty$, then the section integrals are finite for $\nu$-a.e. $y$, and the Fubini identity holds. In the signed or complex case, the section-integral function is defined on that full-measure set and may be assigned any fixed finite value on the exceptional null set to obtain a measurable version.

For nonnegative functions, the inner integral as a function of the parameter is measurable directly by Tonelli, and it may equal $+\infty$ at some parameters. For signed or complex functions, one must handle measurability without presupposing Fubini’s theorem; the decomposition below does so using Tonelli alone. In either case, a finite inner integral at each parameter does not by itself ensure that the parameter integral is finite.

Why the Integrable Case Has a Measurable Version

We record the reduction that permits the Fubini exchange for real- and complex-valued functions. The only product-integration result used in its proof is Tonelli’s theorem for nonnegative functions, established earlier in the course.

Proposition: Let $(X,\mathcal M,\mu)$ and $(Y,\mathcal N,\nu)$ be sigma-finite measure spaces, and let $f:X\times Y\to\mathbb C$ be measurable with $\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty$. Then $y\mapsto\int_X f(x,y)\,d\mu(x)$ is defined and finite for $\nu$-a.e. $y$, has a measurable version, and $$ \int_Y\left(\int_X f(x,y)\,d\mu(x)\right)d\nu(y)=\int_{X\times Y}f\,d(\mu\times\nu). $$ Both sides are finite complex numbers; the section integral on the left is interpreted using the measurable version.

Proof. Write $f=u+iv$, where $u$ and $v$ are real-valued, and write $u=u^+-u^-$ and $v=v^+-v^-$, with all four positive parts nonnegative and measurable. Each is bounded above by $|f|$, so its product integral is finite. By Tonelli, the four functions $$ U_\pm(y)=\int_X u^\pm(x,y)\,d\mu(x),\qquad V_\pm(y)=\int_X v^\pm(x,y)\,d\mu(x) $$ are measurable and have finite integrals over $Y$. A nonnegative measurable function with finite integral is finite a.e.: indeed, if it were infinite on a set of positive measure, its integral would be infinite. Thus all four section integrals are finite outside a measurable $\nu$-null set $N$, the union of their exceptional sets. On $Y\setminus N$, define $G=(U_+-U_-)+i(V_+-V_-)$, which equals the section integral of $f$. Define $G=0$ on $N$. This is measurable, because each $U_\pm,V_\pm$ is measurable and $N\in\mathcal N$. By Tonelli applied separately to the four nonnegative parts, their integrals over $Y$ equal their corresponding product integrals. Subtracting the finite real integrals and combining real and imaginary parts gives $\int_Y G\,d\nu=\int_{X\times Y}f\,d(\mu\times\nu)$. The absolute value of $G$ is bounded a.e. by $\int_X|f(x,y)|\,d\mu(x)$, whose integral is $\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty$ by Tonelli. Hence the parameter integral is integrable and the stated equality is meaningful. $\square$

To exchange the variables, apply the same argument with $X$ and $Y$ interchanged. For a nonnegative function, Tonelli applies directly in either order. These rules justify the examples that follow; each example specifies which condition is being used.

Indicator Functions and Moving Regions

Worked Example: Integrating Over a Triangle

On $(\mathbb R,\mathcal L,m)\times(\mathbb R,\mathcal L,m)$, let $A=\{(s,t):0<t<s<1\}$ and $f=\chi_A$. Find the parameter integral in $t$ and compute the total integral by both orders. The set is Borel, hence product-measurable, and $f$ is nonnegative, so Tonelli applies.

Fix $s$. If $0<s<1$, the section $\{t:0<t<s\}$ has Lebesgue measure $s$. If $s\notin(0,1)$, it is empty. Therefore the parameter integral is $G(s)=s\chi_{(0,1)}(s)$, including $G(s)=0$ at $s=0$ and $s=1$. Integrating it gives

\int_{\mathbb R}G(s)\,dm(s)=\int_0^1s\,ds=\left[\frac{s^2}{2}\right]_0^1=\frac12.

For the reverse order, fix $t$. When $0<t<1$, the allowed values are $t<s<1$, an interval of measure $1-t$; outside that range of $t$, the section is empty. Thus

\int_{\mathbb R}\left(\int_{\mathbb R}\chi_A(s,t)\,dm(s)\right)dm(t) =\int_0^1(1-t)\,dt =\left[t-\frac{t^2}{2}\right]_0^1 =\frac12.

Check: both parameterizations give $1/2$. The strict inequalities defining $A$ exclude its boundary, but that boundary has two-dimensional Lebesgue measure zero and does not change either integral.

Worked Example: An Exponential Weight on a Moving Interval

Evaluate $$ I=\int_0^\infty e^{-x}\left(\int_0^x 1\,dt\right)dx $$ by reversing the order. The integrand on the product space is $e^{-x}\chi_{\{(x,t):0<t<x\}}$, a nonnegative measurable function on $\mathbb R^2$. Tonelli applies, with infinite values permitted until the computation establishes finiteness.

First, $\int_0^x1\,dt=x$ for $x>0$, so integration in the displayed order gives $$ I=\int_0^\infty xe^{-x}\,dx =\left[-xe^{-x}\right]_0^\infty+\int_0^\infty e^{-x}\,dx =0+1=1. $$ Here $xe^{-x}\to0$ as $x\to\infty$, and $\int_0^\infty e^{-x}\,dx=[-e^{-x}]_0^\infty=1$.

For the reverse order, fix $t>0$. The condition $0<t<x$ leaves $x\in(t,\infty)$, and $\int_t^\infty e^{-x}\,dx=e^{-t}$. Hence $$ I=\int_0^\infty e^{-t}\,dt=[-e^{-t}]_0^\infty=1. $$ Check: the first order reduces to $\int_0^\infty xe^{-x}\,dx=1$, and the reversed order reduces to $\int_0^\infty e^{-t}\,dt=1$. Their agreement is guaranteed by Tonelli, and the explicit values show that the common integral is finite.

Exponential Kernels and a Logarithmic Integral

Worked Example: Reversing an Exponential-Kernel Integral

Evaluate $$ J=\int_0^1\int_0^\infty e^{-(x+1)y}\,dy\,dx. $$ The function is nonnegative and measurable on $(0,1)\times(0,\infty)$, with Lebesgue measure in each variable, so Tonelli permits either order even before finiteness is known.

For each $x\in(0,1)$, $x+1>0$, and direct integration gives $$ \int_0^\infty e^{-(x+1)y}\,dy =\left[-\frac{e^{-(x+1)y}}{x+1}\right]_0^\infty =\frac{1}{x+1}. $$ Therefore $$ J=\int_0^1\frac{1}{x+1}\,dx =[\ln(x+1)]_0^1=\ln 2. $$

Now reverse the order. For each $y>0$, $$ \int_0^1e^{-(x+1)y}\,dx =e^{-y}\left[-\frac{e^{-xy}}{y}\right]_0^1 =\frac{e^{-y}(1-e^{-y})}{y} =\frac{e^{-y}-e^{-2y}}{y}. $$ Consequently Tonelli gives the exact identity $$ \int_0^\infty\frac{e^{-y}-e^{-2y}}{y}\,dy=\ln 2. $$ The integrand is nonnegative because $e^{-y}\ge e^{-2y}$ for $y>0$; its integral is finite because it equals the already computed $J$. Check: evaluating in the original order gives $\ln 2$, while reversing the order yields the displayed exponential-kernel integral, so that integral also equals $\ln 2$.

Check the Hypotheses Before Exchanging

Tonelli and Fubini do not have interchangeable hypotheses. Tonelli applies to nonnegative measurable functions and allows the common value $+\infty$. Fubini applies to real- or complex-valued measurable functions when the integral of the absolute value is finite; in that case the section integrals exist and are finite for almost every parameter, and the common value is finite. In applications, verify product measurability as well as the relevant sign or integrability condition.

Common pitfall: finite inner integrals need not be integrable in the parameter. On $(0,1)\times(0,\infty)$ with product Lebesgue measure, set $f(x,t)=e^{-tx}$. It is nonnegative and measurable. For every $x\in(0,1)$, the inner integral in $t$ is finite, but $$ \int_0^\infty e^{-tx}\,dt=\frac1x,\qquad \int_0^1\frac1x\,dx=+\infty. $$ The last integral diverges, since $\int_\varepsilon^1 dx/x=-\ln\varepsilon\to+\infty$ as $\varepsilon\downarrow0$. Thus it would be wrong to infer finiteness of the double integral merely from finiteness of each inner integral.

Reversing the order confirms the same conclusion rather than producing a finite answer. For $t>0$, $$ \int_0^1e^{-tx}\,dx=\frac{1-e^{-t}}{t}. $$ For $t\ge1$, this is at least $(1-e^{-1})/t$, since $e^{-t}\le e^{-1}$. The integral of this lower bound over $[1,\infty)$ diverges, as $\int_1^R dt/t=\ln R\to+\infty$. Tonelli therefore gives $+\infty$ in both orders. This example satisfies Tonelli’s nonnegativity hypothesis, but not the finite-integral conclusion one would need in order to apply Fubini.

For signed or complex parameter integrals, the safe check is $\int|f|<\infty$ on the product space, not merely the existence of individual section integrals. The proposition above explains why: absolute integrability controls the positive and negative parts (and the real and imaginary parts), making their section integrals finite a.e. and allowing Tonelli to establish measurability and the integral identity without circularly invoking Fubini.

Check Your Understanding

For each question, identify the measure spaces and state whether Tonelli or Fubini justifies the step.

For $f(s,t)=\chi_{\{0<t<s<2\}}(s,t)$, find the parameter integral in $t$ and its integral over $s$.
Explain why Tonelli applies to $e^{-x}\chi_{\{0<t<x\}}$ on the positive quadrant, even before its total integral is known to be finite.
Reverse the order in $\int_0^1\int_0^\infty e^{-(x+2)y}\,dy\,dx$, and compute the value using the original order as a check.
For a complex-valued product-measurable $f$, state the integrability hypothesis that ensures the section integral is finite for almost every parameter.
In the example with $e^{-tx}$ on $(0,1)\times(0,\infty)$, identify precisely why finite inner integrals do not imply a finite double integral.