If we have $(X \times Y, \overline{\Sigma \times \tau}, \lambda)$ a complete measure space with underlying complete spaces $(X, \Sigma, \mu)$ and $(Y, \tau, \nu)$, and $\lambda = \mu \times \nu$, what exactly is a product measure?
The reason I am asking this is that if we have a function $f \in L^{1}(d\lambda)$, then clearly $\int \limits_{X \times Y} |f| \,d\lambda < \infty$. But we can say $f = f\chi_{ \{ (x,y) \mid f(x,y) \neq 0 \} }$, where $\chi_{ \{ (x,y) \mid f(x,y) \neq 0 \} }$ is the characteristic function of $ \{ (x,y) \mid f(x,y) \neq 0 \}$.
Since the two functions are equal, we get that $\int \limits_{X \times Y} |f| \,d\lambda = \int \limits_{X \times Y} |f\chi_{ \{ (x,y) \mid f(x,y) \neq 0 \} }| \,d\lambda = \int \limits_{X \times Y} |f| \,d( \chi_{ \{ (x,y) \mid f(x,y) \neq 0 \} } \,d\lambda)$.
This means that $f \in L^{1}(d( \chi_{ \{ (x,y) \mid f(x,y) \neq 0 \} } \,d\lambda))$.
But my professor said that the measure $\chi_{ \{ (x,y) \mid f(x,y) \neq 0 \} } \,d\lambda$ is a $\sigma$-finite measure, but not necessarily a product measure. What does that mean? Why can we integrate with it over $X \times Y$ if it is not a product measure? Does it possibly mean that the measure is not necessarily defined on the product $\sigma$-algebra? This measure is defined on the product $\sigma$-algebra, so that can't be it…
Note that if $E \in \overline{\Sigma \times \tau}$, then the measure of $E$ with respect to $\chi_{ \{ (x,y) \mid f(x,y) \neq 0 \} } \,d\lambda$ is defined to be $\int \limits_{E} \chi_{ \{ (x,y) \mid f(x,y) \neq 0 \} } \,d\lambda$.
Best Answer
If $(X, \Sigma)$ and $(Y, \tau)$ are measurable spaces, then a measure $\lambda$ on $(X \times Y, \Sigma \times \tau)$ is said to be a product measure if there exist measures $\mu, \nu$ on $(X, \Sigma)$, $(Y, \tau)$ respectively, such that $\lambda = \mu \times \nu$. That is, for each $A \in \Sigma$, $B \in \tau$, we have $\lambda(A \times B) = \mu(A) \nu(B)$.
Not every measure on $(X \times Y, \Sigma \times \tau)$ is a product measure. For a simple example, take $(X, \Sigma) = (Y, \tau) = (\{a,b\}, \mathcal{P}(\{a,b\}))$ to be sets with two points, and define, for instance, $\lambda(\{(a,a)\}) = \lambda(\{(b,b)\}) = 1$ and $\lambda(\{(a,b)\}) = \lambda(\{(b,a)\}) = 0$, defined for other sets by additivity. Suppose $\lambda = \mu \times \nu$. Since $\{(a,b)\} = \{a\} \times \{b\}$, we must have either $\mu(\{a\})=0$, contradicting $\lambda(\{(a,a)\}) = \mu(\{a\}) \nu(\{a\}) = 1$, or $\nu(\{b\})=0$, contradicting $\lambda(\{(b,b)\}) = \mu(\{b\}) \nu(\{b\}) = 1$. Either way we reach a contradiction, so $\lambda$ is not a product measure.