Mixed Counting-And-Lebesgue Product Integrals

From a Product Integral to a Sum of Integrals

A product space with one countable coordinate and one real-valued coordinate has a useful special feature: integration in the countable coordinate is summation. Thus a product integral can often be evaluated by integrating one real-variable slice at a time, then adding the results.

This translation is useful for discrete-continuous models. For instance, a random pair may have a countable-valued first coordinate and a real-valued second coordinate. Its joint density, with respect to counting measure and Lebesgue measure, is a function on a product space. Integrating that function means summing its one-dimensional integrals over the discrete values.

The Mixed Product Space

Let \(E\) be a countable set, equipped with its power set and counting measure \(\#\), and let \(\mathbb R\) be equipped with Lebesgue measure \(\lambda\). Counting measure assigns mass one to each point, so the integral of a nonnegative function \(a:E\to[0,\infty]\) is \(\sum_{s\in E}a(s)\). On \(E\times\mathbb R\), use the product measure \(\#\times\lambda\).

Definition: For a measurable function \(f:E\times\mathbb R\to[0,\infty]\), the slice at \(s\in E\) is the function \(f_s:\mathbb R\to[0,\infty]\) given by \(f_s(x)=f(s,x)\). The mixed product integral is \(\int_{E\times\mathbb R} f\,d(\#\times\lambda)\).

Every slice of a product-measurable function is Lebesgue measurable. When \(E\) is countable with its power set, the product-measurable structure can also be understood slice by slice: a function on \(E\times\mathbb R\) is measurable precisely when each of its real-variable slices is measurable.

The central formula is Tonelli’s theorem specialized to this product. Because counting measure on a countable set is sigma-finite and Lebesgue measure is sigma-finite, the theorem applies. The formula permits infinite values when the integrand is nonnegative.

Nonnegative Integrands

Theorem: If \(f:E\times\mathbb R\to[0,\infty]\) is measurable, then

\int_{E\times\mathbb R} f(s,x)\,d(\#\times\lambda)(s,x) = \sum_{s\in E}\int_{\mathbb R} f(s,x)\,d\lambda(x) = \int_{\mathbb R}\sum_{s\in E}f(s,x)\,d\lambda(x).

Proof. Apply Tonelli’s theorem to the two sigma-finite measure spaces \((E,\#)\) and \((\mathbb R,\lambda)\). Its iterated-integral formulas give the first equality and allow the order of integration to be reversed, giving the second. Integration with respect to counting measure is summation, which yields the displayed expressions. All quantities are well-defined in \([0,\infty]\), so no finiteness assumption is required.

\(\square\)

In practical use, the first equality is often the most direct: fix each discrete value, integrate over the real coordinate, and sum. The second equality can be convenient when, for each fixed \(x\), the sum over \(s\) has a simple form.

Worked Example: Slices Partition the Positive Half-Line

Take \(E=\mathbb N_0\) and define \(f(n,x)=e^{-x}\mathbf 1_{[n,n+1)}(x)\), where \(\mathbf 1_A\) is the indicator of the set \(A\). This is nonnegative and measurable. For each \(n\), its slice integral is

\int_{\mathbb R}f(n,x)\,dx = \int_n^{n+1}e^{-x}\,dx = e^{-n}(1-e^{-1}).

Summing the slice integrals gives

\sum_{n=0}^\infty e^{-n}(1-e^{-1}) = (1-e^{-1})\frac{1}{1-e^{-1}} = 1.

Indeed, the intervals \([n,n+1)\) partition \([0,\infty)\), and the product integral is also \(\int_0^\infty e^{-x}\,dx=1\). This verifies the mixed-integral formula in a case where the discrete index selects disjoint real-variable slices.

Discrete-Continuous Joint Densities

If \(E\) is the set of possible values of a discrete coordinate, a nonnegative measurable function \(p:E\times\mathbb R\to[0,\infty]\) is a joint density with respect to \(\#\times\lambda\) when its product integral is one. For a discrete-continuous probability model, the mass assigned to a discrete value is not included automatically in counting measure: it must appear in the function \(p\).

Worked Example: Normalizing a Joint Density and Finding an Event Probability

On \(\mathbb N_0\times\mathbb R\), define

p(n,x)=2^{-(n+1)}e^{-x}\mathbf 1_{(0,\infty)}(x).

Every slice is nonnegative, and its integral is \(2^{-(n+1)}\), because \(\int_0^\infty e^{-x}\,dx=1\). Therefore

\sum_{n=0}^\infty\int_{\mathbb R}p(n,x)\,dx = \sum_{n=0}^\infty 2^{-(n+1)} = 1.

Thus \(p\) is a probability density. To find the probability of the event that the discrete coordinate is even and the real coordinate exceeds \(1\), sum over \(n=2k\), \(k\in\mathbb N_0\), and integrate over \(x>1\):

\begin{aligned} \sum_{k=0}^\infty\int_1^\infty p(2k,x)\,dx &= \sum_{k=0}^\infty 2^{-(2k+1)}e^{-1}\\ &= e^{-1}\frac{1/2}{1-1/4} = \frac{2}{3e}. \end{aligned}

The factor \(e^{-1}\) is the integral of \(e^{-x}\) over \((1,\infty)\); the geometric series accounts for all even discrete values.

Integrable Signed and Complex Integrands

For a signed or complex function, cancellation can make an iterated expression appear finite even when the product integral is not defined as a finite integral. The safe condition is absolute integrability: the integral of the absolute value must be finite. In the mixed setting, Tonelli turns that condition into a sum of ordinary slice integrals.

Theorem: Suppose \(f:E\times\mathbb R\to\mathbb C\) is measurable and

\int_{E\times\mathbb R}|f(s,x)|\,d(\#\times\lambda)(s,x) = \sum_{s\in E}\int_{\mathbb R}|f(s,x)|\,d\lambda(x) < \infty.

Then \(f\) is integrable, the sum \(\sum_{s\in E}\int_{\mathbb R}f(s,x)\,d\lambda(x)\) converges absolutely, and

\int_{E\times\mathbb R} f\,d(\#\times\lambda) = \sum_{s\in E}\int_{\mathbb R}f(s,x)\,d\lambda(x) = \int_{\mathbb R}\sum_{s\in E}f(s,x)\,d\lambda(x).

Proof. Tonelli’s theorem applied to \(|f|\) gives the equality in the hypothesis. In particular, \(\int |f|<\infty\), so \(f\) is integrable. For every \(s\), the integral of \(f(s,\cdot)\) exists, and

\sum_{s\in E}\left|\int_{\mathbb R}f(s,x)\,d\lambda(x)\right| \leq \sum_{s\in E}\int_{\mathbb R}|f(s,x)|\,d\lambda(x) < \infty.

For real-valued \(f\), apply Tonelli separately to \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). Their integrals are finite because each is bounded above by \(|f|\). Subtracting their nonnegative integral formulas gives the asserted identities. For complex-valued \(f\), apply the real-valued result to the real and imaginary parts; both are integrable because their absolute values are bounded by \(|f|\). This proves the formulas, including the iterated integral in the reverse order.

\(\square\)

The theorem also explains how to check the hypothesis without first integrating over the product space: calculate \(\int_{\mathbb R}|f(s,x)|\,dx\) for each \(s\), then sum those nonnegative quantities. Finite total is the relevant guarantee; conditional cancellation in a series is not a substitute.

Worked Example: An Absolutely Integrable Signed Function

Let \(E=\mathbb N_0\) and set

f(n,x)=(-1)^n2^{-(n+1)}(1+x)e^{-x}\mathbf 1_{[0,\infty)}(x).

First, integration by parts gives \(\int_0^\infty xe^{-x}\,dx=1\), and direct integration gives \(\int_0^\infty e^{-x}\,dx=1\). Hence \(\int_0^\infty(1+x)e^{-x}\,dx=2\). The absolute integral is therefore

\sum_{n=0}^\infty\int_{\mathbb R}|f(n,x)|\,dx = 2\sum_{n=0}^\infty 2^{-(n+1)} = 2.

Absolute integrability permits the mixed-integral formula. The slice integrals yield

\sum_{n=0}^\infty\int_{\mathbb R}f(n,x)\,dx = 2\sum_{n=0}^\infty(-1)^n2^{-(n+1)} = 2\left(\frac{1/2}{1+1/2}\right) = \frac{2}{3}.

Summing first gives the same result: for every \(x\geq0\), the geometric series over \(n\) equals \(1/3\), so the resulting integral is \(\frac13\int_0^\infty(1+x)e^{-x}\,dx=2/3\). Both orders agree, and the absolute-integral calculation justifies the exchange.

Reading the Formula Correctly

Two conventions are worth keeping distinct. First, counting measure assigns unit mass to every discrete point. If instead the discrete coordinate has weights \(q_s\), integration with respect to that weighted measure produces \(\sum_{s\in E}q_s\int f(s,x)\,dx\); those weights must not be silently omitted or counted twice. Second, for nonnegative functions the sum and integral may both be infinite, while for signed or complex functions finite absolute integrability is what licenses the finite-valued Fubini formulas.

These rules turn many calculations on mixed spaces into familiar one-dimensional integrals followed by a series. They also provide a practical verification method: establish measurability, identify the measure used for the discrete coordinate, and check nonnegativity or absolute integrability before changing the order of operations.

Check Your Understanding

Use the slice formulas to answer these questions.

What does integration with respect to counting measure do to a nonnegative function on a countable set?
For a nonnegative measurable function on \(E\times\mathbb R\), may either iterated integral be infinite?
What condition on a signed or complex function ensures that the sum of its slice integrals converges absolutely?
In a density with respect to counting measure times Lebesgue measure, where must the discrete probability weights appear?
For the signed example, what is the value of the sum of the absolute slice integrals?