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Let $(X,\mathscr{A},\mu), (Y,\mathscr{B},\nu)$ be two measure spaces, then we have the product measurable space $(X\times Y, \mathscr{A}\times\mathscr{B})$ where $\mathscr{A}\times\mathscr{B}$ is the $\sigma$-algebra generated by $\{A\times B, A\in \mathscr{A}, B\in \mathscr{B}\}$. Applying Caratheodory extension we can always construct a measure $m$ on $\mathscr{A}\times\mathscr{B}$ so that $m(A\times B)=\mu(A)\nu(B)$. Further, if $\mu$ and $\nu$ are $\sigma$-finite, by monotone class theorem such a measure $m$ is uniquely determined. Now my question is:

When $\mu$ and $\nu$ are not $\sigma$-finite, is product measure still unique?

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Uniqueness does *not* hold in general.

Besides the usual product measure of two measure spaces $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$, there is a different version of the product measure, called the *complete locally determined product*. For products of non-$\sigma$-finite measure spaces it has many desirable properties that the usual product measure lacks. In his *Measure Theory*, Fremlin goes so far as to call the usual product measure the *primitive product measure*, see chapter 5 of volume 2 for more details on the construction and its basic properties.

**What follows below is taken more or less directly from Fremlin’s exposition.**

The construction of the complete locally determined version of the product measure is as follows: Let $\pi = \mu \times \nu$ be the usual product measure on the *full* $\sigma$-algebra $\mathfrak{P} = \mathfrak{M} \otimes \mathfrak{N}$ obtained from performing the Carathéodory construction on the measurable rectangles. It seems more natural to do this than to restrict to the $\sigma$-algebra $\mathfrak{M} \times \mathfrak{N} \subset \mathfrak{M} \otimes \mathfrak{N}$ generated by the measurable rectangles from the beginning. The *complete locally determined product measure* $p$ is an *“inner regularization”* of the usual product measure $\pi$. Its definition is that for $P \in \mathfrak{P}$ one puts

$$p(P) = \sup{\{\pi(P \cap (M \times N))\,:\,\mu(M) \lt \infty, \nu(N) \lt \infty\}}.$$

If you’re willing to believe that $p$ is a measure, you can skip the next section of this answer.

Let us check that $p$ is indeed a measure: Clearly, $p(\emptyset) = 0$. If $P_n \in \mathfrak{P}$ is a sequence of pairwise disjoint sets, we can estimate for every pair $M \in \mathfrak{M},N \in \mathfrak{N}$ with $\mu(M) \lt \infty$ and $\nu(N) \lt \infty$ that

$$

\pi \left( \bigcup_{n=1}^{\infty} P_n \cap (M \times N)\right) = \sum_{n=1}^{\infty}\; \pi\left(P_n \cap (M \times N)\right) \leq \sum_{n=1}^{\infty}\;p(P_n),

$$

hence

$p\left(\bigcup_{n=1}^{\infty}P_n\right) \leq \sum_{n=1}^{\infty}\;p(P_n),$ so $p$ is $\sigma$-subadditive. To prove $\sigma$-additivity, note that we can choose for every $t \lt \sum p(P_n)$ a large enough $k$ such that there are

$t_1 \lt p(P_1), \ldots, t_k \lt p(P_k)$ with $t \lt t_1 + \cdots + t_k$. Then we can choose $M_i \in \mathfrak{M},N_i \in \mathfrak{N}$ with

$\mu(M_i)\lt \infty$ and $\nu(N_i) \lt \infty$ such that $t_i \leq \pi(P_i \cap (M_i \times N_i))$. Let $M = M_1 \cup \cdots \cup M_k$, $N = N_1 \cup \cdots \cup N_k$. Then, by definition, and the fact that $\mu(M) \lt \infty$ and $\nu(N)\lt \infty$, we have

$$

p\left(\bigcup_{n=1}^\infty P_n\right) \geq

\pi\left(\bigcup_{n=1}^\infty P_n \cap (M \times N)\right) \geq

\sum_{n = 1}^k \; \pi (P_n \cap (M \times N)) \geq

\sum_{n=1}^k \; \pi(P_n \cap (M_n \times N_n)) \gt t.

$$

Thus $t \lt p(\bigcup P_n) \leq \sum p(P_n)$ for every $t < \sum p(P_n)$. It follows that $p$ is $\sigma$-additive on disjoint sequences sets in $\mathfrak{P}$, hence $p$ is a measure on $\mathfrak{P}$.

**Exercise 1:** It is clear that $p(M \times N) = \mu(M)\,\nu(N)$ for $M$ and $N$ of finite measure. To ensure this equality for all measurable sets $M$ and $N$, we need to suppose that $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$ are *semi-finite* in the sense that every set of infinite measure contains a measurable set of finite positive measure. This implies that $\mu(M) = \sup{\{\mu(E)\,:\,E \subset M, \mu(E) \lt \infty\}}$ and similarly for $\nu$. Using this, check that the equality $p(M \times N) = \mu(M)\, \nu(N)$ holds for all $M \in \mathfrak{M}$ and $N \in \mathfrak{N}$.

**Exercise 2:** Let $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$ be $\sigma$-finite spaces. Prove that $p = \pi$.

**Exercise 3:** Let $(X,\mathfrak{M},\mu)$ and $(Y,\mathfrak{N},\nu)$ be arbitrary measure spaces. If $\lambda$ is any measure on $\mathfrak{P}$ satisfying $\lambda(M \times N) = \mu(M)\nu(N)$ then $p(P) \leq \lambda(P) \leq \pi(P)$.

We are finally ready to concoct our counterexample—most of the basic counterexamples in connection with product measures are variations of this theme.

Let $(X,\mathfrak{M},\mu) = ([0,1],\Sigma,\lambda)$ be the unit interval with Lebesgue measure. Let $(Y,\mathfrak{N},\nu) = ([0,1],\mathcal{P}([0,1]),\#)$ be the unit interval equipped with counting measure. Let $\pi$ be the usual product measure and $p$ its locally determined version. Since both $\mu$ and $\nu$ are semi-finite, we have by exercise 1 above that $p(M \times N) = \mu(M) \nu(N)$ for all measurable $M \subset X$ and $N \subset Y$.

Notice that the diagonal $\Delta = \{(x,x) \,:\,x \in [0,1]\}$ is in the $\sigma$-algebra $\mathfrak{M \times N}$ generated by the measurable rectangles, because

$$\Delta = \bigcap_{n =1}^{\infty}\; \bigcup_{k=0}^{n-1}\; \left[\frac{k}{n},\frac{k+1}{n}\right] \times \left[\frac{k}{n},\frac{k+1}{n}\right].$$

**Exercise 4:** $\pi(\Delta) = \infty$.

**Exercise 5:** $p(\Delta) = 0$.

Thus, $\pi$ and $p$ are distinct, while agreeing on the measurable rectangles themselves. Since $\Delta \in \mathfrak{M} \times \mathfrak{N}$ this also holds for the restrictions of $\pi$ and $p$ from $\mathfrak{M \otimes N}$ to $\mathfrak{M \times N}$.

I give more simple example. Let $\mu =\lambda \times \#$ be usual product of $\lambda$ and $\#$. Let $\mu_1$ be defined as follows: $$\mu_1(X)=\sum_{y \in [0,1]}\lambda_y(([0,1]\times \{y\})\cap X)$$for all Lebesgue measurable subset $X$ of $[0,1]\times [0,1]$, where $\lambda_y$ is Lebesgue measure defined on $[0,1]\times \{y\}$. Then measures $\mu$ and $\mu_1$ are agreeing on measurable rectangles but $\mu(\{(x,x):x \in [0,1]\})=+\infty$ and $\mu_1( (\{(x,x):x \in [0,1]\})=0$.

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