Question

Let $X=Y=N,\mathcal{M}=\mathcal{N}=\mathcal{P}(\mathrm{N}),\mu =\nu =$ counting measure. Define $f(m,n)=1$ if $m=n,f(m,n)=-1$ if $m=n+1$, and $f(m,n)=0$ otherwise. Then $\int |f|d(\mu \times \nu )=\mathrm{\infty }$, and $\iint fd\mu d\nu$ and $\iint fd\nu d\mu$ exist and are unequal.

Short answer

$\int |f|d(\mu \times \nu )=\mathrm{\infty }$, $\iint fd\mu d\nu =0$, $\iint fd\nu d\mu =-1$.

Step-by-step solution

Understanding the Function

The function $f(m,n)$ is given as follows: $f(m,n)=1$ if $m=n$, $f(m,n)=-1$ if $m=n+1$, and $f(m,n)=0$ otherwise. This means the function is non-zero only when $m$ is either equal to $n$ or equal to $n+1$.

Setting Up the Integral

We need to evaluate $\int |f|d(\mu \times u)$ where $\mu \times u$ is the product measure of the counting measure. Here, $|f(m,n)|=1$ if $m=n$ or $m=n+1$, and $0$ otherwise.

Evaluating $\int |f|d(\mu \times \nu )$

The integral $\int |f|d(\mu \times u)$ involves summing over all pairs $(m,n)$. Since there are infinitely many pairs $(n,n)$ and $(n+1,n)$ both contributing $1$ to $|f|$, the result is indeed infinite: $\int |f|d(\mu \times u)=\mathrm{\infty }$.

Evaluating $\iint fd\mu d\nu$

To find $\iint fd\mu du$, integrate $f(m,n)$ with respect to $u$ for each fixed $m$, then integrate the result with $\mu$. For each fixed $m=n$ and $m=n+1$, the result of the inner integral is zero since contributions of $1$ and $-1$ cancel out when summing over $n$, making $\iint fd\mu du=0$.

Evaluating $\iint fd\nu d\mu$

To find $\iint fdud\mu$, integrate $f(m,n)$ with respect to $\mu$ for each fixed $n$, then integrate the result with $u$. Here, each term $f(m,n)$ contributes $1$ when $m=n$ and $-1$ when $m=n+1$. However, for each $n$, there is no equal sum of positive and negative since $m=n+2,n+3,\dots$ keep contributing zero. The summation is zero for all terms, but the contributions don't cancel as in the previous case resulting in $\iint fdud\mu =-1$.

Key concepts

Counting Measure

Counting measure is a type of measure used primarily for discrete sets, such as the set of natural numbers, integers, or any finite set. It is the simplest form of a measure, where the size of a subset is simply the number of elements it contains. For example, if we have a set with elements $\{1,2,3\}$, the counting measure would be $3$ because there are three elements.

This type of measure is particularly useful in fields like number theory and combinatorics where discrete scenarios are common. In the context of the exercise, both $X$ and $Y$ are the set of natural numbers with the power set $\mathcal{P}(N)$ serving as their sigma-algebra. The measure $\mu =u=$ counting measure applies to all subsets of $N$ and assigns a value equal to the cardinality of the subset.

Product Measure

In measure theory, a product measure is constructed on a product space $X\times Y$ from two measure spaces $(X,\mathcal{M},\mu )$ and $(Y,\mathcal{N},u)$. It helps evaluate measures over product spaces and combines individual measures from each dimension.

The product measure $\mu \times u$ of our specific exercise is formed by multiplying the counting measures from two identical spaces $N$. Thus, it considers pairs $(m,n)$ where each pair can be considered separately through its marginal measures. This setup allows analysis of functions defined over $N\times N$, such as our function $f(m,n)$.

• Each slice $(m,n)$ is evaluated by counting measure separately.
• In our example, this precisely means the measure is connected with how many discrete points, such as $(n,n)$ and $(n+1,n)$, exist.

Fubini's Theorem

Fubini's Theorem is an essential tool in calculus that gives us a method to evaluate a double integral as iterated, single integrations. It allows us to switch the order of integration when dealing with integrals over product spaces.

In the example, Fubini's theorem was applied but with different results depending on the order of integration: $\iint fd\mu du$ versus $\iint fdud\mu$. While the theorem provides the theoretical ground to swap integrals, the non-absolutely convergence of $|f|$ results in these integrals not being equal.

• When integrated by $\mu$ first, and then by $u$, the positive and negative contributions of $f$ cancel out perfectly, resulting in zero.
• Meanwhile, using the converse order, does not fully cancel all contributions, hence resulting in $-1$.

Double Integral

A double integral is an integral over a two-dimensional area and is a way of summing a function over an area in the plane. In measure theory, integrals of this kind are used when working with functions defined on product spaces.

In our exercise, the double integral $\int |f|d(\mu \times u)$ essentially evaluates the total absolute value of our given function over all pairs $(m,n)$. The result was infinite because the function was non-zero at infinitely many points, even though they only contributed 1 at those positions. This means that the absolute sum never converged, hence yielding an infinite result.

• Double integrals let us explore dimensions that single integrals can't.
• They empower us to handle more complex multi-dimensional problems like those in probability and physics.
• In this exercise, it revealed the divergence when using traditional methods over spaces with infinitely many elements.