Why the product of measure is a tensor product?

Question

Let $\mu$ and $\nu$ two measure. What is the justification to say that the measure product is a tensor product ? i.e. why $\mu\times \nu$ is written $\mu\otimes \nu$ ? (Of course, beside the fact that is by definition or it's just a notation). Our teacher told us that the reason that we write it as a tensor product it's because it behave as a tensor product). But I don't understand this (I'm not very confortable with tensor product).

Answer 1

In my opinion, this should simply be called the product of the two measurable spaces. In fact, it satisfies the corresponding universal property (When the category treated is the category of measure spaces $(X,\mu)$ with the natural morphisms.

However, a different way of viewing the operations is to think of our category as a monoidal category, where the binary operation is $(X,\mu)\otimes(Y,\nu)\rightarrow(X\times Y,\mu\otimes \nu)$, Every category with a product operation has a monoidal structure and in the case of vector spaces, that operation is the tensor product.

Answer 2

If $f$ is a real-valued function of the variable $x\in X$, and $g$ is a function of $y\in Y$, then the function of two variables $$ (x, y)\in X\times Y \mapsto f(x)g(y)$$ is called tensor product of $f$ and $g$ and denoted by $f\otimes g$. The tensor product of measures extends this definition; indeed, if $\mu$ and $\nu$ have densities $f$ and $g$ respectively, then $$ \mu\otimes \nu = f\otimes g\, dxdy.$$ (Here, of course, $x$ and $y$ are interpreted as real variables.)

Answer 3

For every measure space $(\Omega, \Sigma, m)$ and every $f=f(\omega) \in L^1(\Omega, \Sigma, m)$, let's define $m(f(\omega))$ as follows: \begin{equation} m(f(\omega)):=\int_{\Omega}f(\omega)dm(\omega). \end{equation} If $f_1(x),f_2(x),\dots f_n(x)\in L^1(X,\Sigma,\mu)$ and $g_1(y), g_2(y), \dots g_n(y) \in L^1(Y,\Sigma_2,\nu)$, then: \begin{align} (\mu \otimes \nu) \Big(\sum_{k=1}^n f_k(x)g_k(y)\Big) &= \sum_{k=1}^n (\mu \otimes \nu) ( f_k(x)g_k(y))\\ &= \sum_{k=1}^n \mu(f_k(x)) \nu(g_k(y)) \end{align} where the latter equality is true thanks to Fubini's theorem.

If you look at the construction of the tensor product of linear mappings and compare it with these equations, you can see they share the same pattern.