Fubini's Theorem for Integrable Functions

From Product Integrals to Section Integrals

An integral on a product space can describe the total contribution of a function depending on two variables. In applications, however, it is often useful to hold one variable fixed, integrate over the other, and then integrate the result. Fubini’s theorem gives precise conditions under which this procedure is valid.

The key condition is absolute integrability. It ensures not only that the product-space integral is finite, but also that almost every section is integrable and that either order of integration gives the same value. The qualification “almost every” matters: a function can be integrable on the product space while having an exceptional section that is not integrable.

Setting and Definition

Let \((X,\mathcal A,\mu)\) and \((Y,\mathcal B,\nu)\) be measure spaces, and suppose both measures are sigma-finite. Write \(\mu\otimes\nu\) for the product measure on \(\mathcal A\otimes\mathcal B\). For a product-measurable function \(f:X\times Y\to\mathbb R\), define its sections by \(f_x(y)=f(x,y)\) and \(f^y(x)=f(x,y)\). The section measurability result for product-measurable functions ensures that these sections are measurable.

Definition: A real-valued measurable function \(f\) on \(X\times Y\) is integrable if the integral of its absolute value with respect to the product measure is finite. In symbols,

\int_{X\times Y}|f|\,d(\mu\otimes\nu)<\infty.

The earlier Tonelli theorem for nonnegative measurable functions will be the main tool. In the sigma-finite setting, it says that the section integrals of a nonnegative product-measurable function are measurable and that its product integral equals either iterated integral, including when the values are infinite. Applying this result to \(|f|\) will control the sections of an integrable function.

Fubini’s Theorem

Theorem: Let \((X,\mathcal A,\mu)\) and \((Y,\mathcal B,\nu)\) be sigma-finite measure spaces, and let \(f:X\times Y\to\mathbb R\) be product-measurable and integrable. Then \(f_x\) is integrable for \(\mu\)-almost every \(x\), and \(f^y\) is integrable for \(\nu\)-almost every \(y\). The functions given by the section integrals are measurable after assigning arbitrary finite values on their exceptional null sets. Both iterated integrals exist as finite integrals and satisfy

\int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x) = \int_{X\times Y}f\,d(\mu\otimes\nu) = \int_Y\left(\int_X f(x,y)\,d\mu(x)\right)d\nu(y).

Proof: Apply Tonelli’s theorem to the nonnegative function \(|f|\). The function

h(x)=\int_Y |f(x,y)|\,d\nu(y)

is measurable, and

\int_X h\,d\mu = \int_{X\times Y}|f|\,d(\mu\otimes\nu) < \infty.

A nonnegative measurable function with finite integral is finite almost everywhere. Thus \(h(x)<\infty\) for \(\mu\)-almost every \(x\), which says precisely that \(f_x\) is integrable for almost every \(x\). Applying the same argument in the other order shows that \(f^y\) is integrable for \(\nu\)-almost every \(y\).

To establish the equality of integrals, write \(f=f^+-f^-\), where \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). Both functions are nonnegative and product-measurable, and \(f^+,f^-\leq |f|\). Tonelli’s theorem therefore gives

\int_{X\times Y} f^+\,d(\mu\otimes\nu)<\infty, \qquad \int_{X\times Y} f^-\,d(\mu\otimes\nu)<\infty.

For almost every \(x\), both \(\int_Y f^+(x,y)\,d\nu(y)\) and \(\int_Y f^-(x,y)\,d\nu(y)\) are finite. On this full-measure set, the section integral of \(f\) is their difference. Tonelli says that the two nonnegative section-integral functions are measurable; their difference is consequently measurable where both are finite. Assigning a finite value, such as zero, on the remaining null set gives a measurable version of the section integral.

The integrals of the two nonnegative section-integral functions over \(X\) equal the corresponding product integrals by Tonelli. Subtracting these finite equalities yields

\int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x) = \int_{X\times Y}f^+\,d(\mu\otimes\nu) - \int_{X\times Y}f^-\,d(\mu\otimes\nu) = \int_{X\times Y}f\,d(\mu\otimes\nu).

Repeating the same argument with \(x\) and \(y\) interchanged proves the equality for the other order of integration. Since both positive and negative parts have finite integrals, these iterated integrals are finite.
\(\square\)

A Useful Bound for the Section Integral

Fubini also gives a useful quantitative estimate. It shows that integrating out one variable cannot increase the total absolute integral.

Proposition: Under the hypotheses of Fubini’s theorem, define \(g(x)=\int_Y f(x,y)\,d\nu(y)\) wherever the section is integrable, and set \(g(x)=0\) on the exceptional null set. Then \(g\) is integrable and

\int_X |g(x)|\,d\mu(x) \leq \int_{X\times Y}|f|\,d(\mu\otimes\nu).

Proof: For every \(x\) whose section is integrable, the triangle inequality for the integral gives

|g(x)| = \left|\int_Y f(x,y)\,d\nu(y)\right| \leq \int_Y |f(x,y)|\,d\nu(y) = h(x).

On the exceptional null set, \(g\) was set to zero, so \(|g(x)|\leq h(x)\) there as well. Integrating this pointwise inequality and applying Tonelli’s theorem to \(|f|\) gives

\int_X |g|\,d\mu \leq \int_X h\,d\mu = \int_{X\times Y}|f|\,d(\mu\otimes\nu) < \infty.

Thus \(g\) is integrable and satisfies the claimed bound.
\(\square\)

Worked Examples

Worked Example: A Signed Function on the Unit Square

Let \(X=Y=[0,1]\) with Lebesgue measure and \(f(x,y)=x-y\). Since \(|x-y|\leq 1\) on the unit square, \(f\) is absolutely integrable. For each fixed \(x\),

\int_0^1 (x-y)\,dy = \left[xy-\frac{y^2}{2}\right]_{0}^{1} = x-\frac12.

Integrating this section integral over \(x\) gives

\int_0^1\left(x-\frac12\right)dx = \left[\frac{x^2}{2}-\frac{x}{2}\right]_{0}^{1} = 0.

In the other order, for each fixed \(y\),

\int_0^1 (x-y)\,dx = \left[\frac{x^2}{2}-xy\right]_{0}^{1} = \frac12-y, \qquad \int_0^1\left(\frac12-y\right)dy=0.

Both iterated integrals equal the product-space integral, as Fubini guarantees. The negative and positive contributions cancel in each calculation.

Worked Example: An Integrable Function on an Unbounded Product Space

Take \(X=Y=[0,\infty)\) with Lebesgue measure and let \(f(x,y)=(x-y)e^{-x-y}\). To check absolute integrability, use \(|x-y|\leq x+y\):

\int_0^\infty\int_0^\infty |f(x,y)|\,dy\,dx \leq \int_0^\infty\int_0^\infty (x+y)e^{-x-y}\,dy\,dx = 2.

For the last equality, \(\int_0^\infty e^{-t}\,dt=1\) and \(\int_0^\infty te^{-t}\,dt=1\), the second identity following by integration by parts. Integrating first with respect to \(y\) gives

\int_0^\infty (x-y)e^{-x-y}\,dy = e^{-x}\left(x\int_0^\infty e^{-y}\,dy-\int_0^\infty ye^{-y}\,dy\right) = (x-1)e^{-x}.

Therefore the first iterated integral is

\int_0^\infty (x-1)e^{-x}\,dx = \int_0^\infty xe^{-x}\,dx-\int_0^\infty e^{-x}\,dx = 1-1=0.

Interchanging the roles of \(x\) and \(y\) gives the inner integral \((1-y)e^{-y}\), and hence the other order is \(\int_0^\infty(1-y)e^{-y}\,dy=1-1=0\). Absolute integrability justifies both calculations.

Worked Example: An Exceptional Section

Let \(X=Y=\mathbb R\) with Lebesgue measure, and define \(f(x,y)=1\) when \(x=0\) and \(f(x,y)=0\) otherwise. This is product-measurable. Its support \(\{0\}\times\mathbb R\) has product measure zero: it is the countable union of the sets \(\{0\}\times[-n,n]\), each of product measure zero. Thus \(\int_{\mathbb R^2}|f|\,d(\mu\otimes\nu)=0\).

For \(x\ne 0\), the section \(f_x\) is identically zero and has integral zero. For \(x=0\), the section is identically one on \(\mathbb R\), so its integral is infinite and it is not integrable. This single exceptional value of \(x\) is a null set. In the reverse direction, for every fixed \(y\), the section \(f^y\) is the indicator of the singleton \(\{0\}\), whose Lebesgue integral is zero. The iterated integral in either order is therefore zero, with the section integral in the first order assigned any finite value at \(x=0\).

This example shows why the theorem promises integrable sections almost everywhere rather than everywhere. It also shows why values on an exceptional null set must not be used to argue that every individual section is well behaved.

What the Hypotheses Do

The condition \(\int |f|<\infty\) is what makes the proof work: it lets Tonelli control the absolute values of the sections and ensures that the positive and negative parts both have finite integrals. The theorem does not assert that the same conclusion holds for an arbitrary product-measurable function whose positive and negative contributions are both infinite.

Sigma-finiteness is also part of the stated setting. It is the standard hypothesis under which the product-measure results used here apply. When checking a problem, verify the measure-space hypotheses as well as absolute integrability; the fact that an iterated calculation looks finite is not, by itself, a justification for exchanging the order.

The practical sequence is direct: establish product measurability, verify that the integral of \(|f|\) is finite, and then apply Fubini. The conclusion supplies almost-everywhere section integrability and equality of the two iterated integrals. No claim about an exceptional null set is needed for the value of those integrals.

Check Your Understanding

Use the hypotheses and conclusions of Fubini’s theorem to answer these questions.

Which function is the key one to integrate with Tonelli’s theorem when proving Fubini for an integrable real-valued function?
Why does finiteness of the integral of the section function \(h(x)=\int_Y|f(x,y)|\,d\nu(y)\) imply that almost every section is integrable?
In the exceptional-section example, which section fails to be integrable, and why does this not contradict Fubini’s theorem?
What estimate bounds the integral of the absolute value of a section integral by the product-space integral of \(|f|\)?
Why is absolute integrability relevant when subtracting the integrals of the positive and negative parts of \(f\)?