In several texts you can encounter tensor product or Fubini product of two filters defined as follows: If $p$ is a filter on a set $X$, $q$ be a filter on a set $Y$, then $$p\otimes q=\{A\subseteq X\times Y; \{x; \{y; (x,y)\in A\}\in q\}\in p\}$$ is a filter on $X\times Y$ which is called Fubini product or tensor product of $p$ and $q$.

Some authors define them only for ultrafilters. Some authors work dually with ideals. Quite often we work only with situation that $X=Y$. See, for example, [HS, Definition 11.1], [BS, page 125] or [Far, page 8].1

The definition can be roughly rephrased as: For almost every $x$ we have that almost all $y$'s belong to the $x$-section $A_x=\{y\in Y; (x,y)\in A\}$. Of course, we have to interpret almost every using the corresponding filters $p$ and $q$.

Question. What is the intuition behind this definition? Are there some viewpoints in which this operation arises naturally? Are there some important properties which makes this notion useful tool in some situations?

I was able to think at least of something:

  • It seems that this operation is naturally related to iterated limit w.r.t. filter (ultrafilter). I have tried to expand on this in another post: Tensor product of ultrafilters corresponds to iterated limit.
  • I have seen a video where Andreas Blass explains how this is related to Fubini's theorem and the corresponding measure if we view ultrafilters as finitely additive measures. It is a part of the video from Peter Krautzberger's talk Idempotent Ultrafilters: An Introduction approximately about 45:00.
  • This might also be related to this question: Products of ultrafilters and products of measures. Although the connection as explained there seems a bit unclear to me. (I am not sure what authors of [BS] mean by "the usual definition of the product measure" - it is possible that this is the same explanation as in my previous bullet point, just put in another words.)

From the above three bullet points, I have feeling that I really understand only the first one. Still, I will be grateful if somebody expands on any of them. And certainly if somebody mentions another possibility how one might look at this type of product.


  • [BS] J. L. Bell, A. B. Slomson: Models and Ultraproducts: An Introduction, North-Holland, 1974.
  • [HS] N. Hindman, D. Strauss: Algebra in the Stone-Cech compactification, De Gruyter, 2012, Second edition.
  • [Far] Farah I. Analytic Quotients. Theory of Liftings for Quotients over Analytic Ideals on the Integers, Memoirs of AMS, 2000.

1You may notice that the definition in [BS] reverses order compared to [HS]. As far as I can tell the definition which I have included is more frequent. Both [BS] and [HS] work only with ultrafilters. In [Far] this kind of product is defined dually for ideals. The definition given there is that Fubini product of ideals $\mathcal I$ and $\mathcal J$ is given by $$A\in\mathcal I\times \mathcal J \qquad\Leftrightarrow\qquad \{i; A_i\notin\mathcal I\}\in\mathcal J.$$ If we denote $p$ the dual filter to $\mathcal I$ and $q$ the dual filter to $\mathcal J$, and recall that dual filter contains exactly the complements of sets belonging to the ideal, then this basically says \begin{align*} A\in p\otimes q &\Leftrightarrow X\setminus A\in\mathcal I\times \mathcal J \\ &\Leftrightarrow \{i; (X\setminus A)_i\notin\mathcal I\}\in\mathcal J \\ &\Leftrightarrow \{i; (X\setminus A)_i\in\mathcal I\}\in q \\ &\Leftrightarrow \{i; A_i\in\mathcal p\}\in q \\ \end{align*}

  • In topology, filters help generalize concepts like limits. (See also: nets. For example, the Riemann integral is defined in terms of such a generalized limit.) Here's a topological characterization of the filter product:

    Theorem: If $f:X\times Y \rightarrow T$ is a function into a compact Hausdorff space $T$, and if $p$ and $q$ are ultrafilters on $X$ and $Y$ respectively, then there exists a unique product filter $p\otimes q$ such that:

    $$\lim_{x\rightarrow p}\lim_{y\rightarrow q} f(x,y) = \lim_{\langle x,y\rangle\rightarrow p\otimes q} f(x,y)$$

    in that both limits exist and are equal.

  • In measure theory, filters are (isomorphic with) a special kind of measure: if $F$ is a filter on $X$, then the function $p(A) \equiv \{1\text{ if }A\in F \text{ else } 0 \} $ is a 0-1 valued, finitely-additive measure on $X$. Conversely, if $p$ is any 0-1 valued, finitely-additive measure on $X$, then $\{A : p(A)=1\}$ is a filter.

    Products of measures are defined in a straightforward way: if $\mu$ and $\nu$ are measures on spaces $X$ and $Y$, then $(\mu\times \nu)(A \times B) \equiv \mu(A)\cdot \nu(B)$ is their product measure on the space $X\times Y$ generated by measureable rectangles. It looks like this isomorphism between 0-1 valued finitely-additive measures and filters preserves products: the filter corresponding to the product of two measures is the product filter for the corresponding two filters.

    $(\mu\times \nu)(A\times B) = 1 \iff [\mu(A) = 1 \wedge \mu(B) = 1] \iff [A \in F \wedge B \in G] \iff [A\times B] \in (F\otimes G)$.

  • $\begingroup$ Product of measures is commutative while tensor product of filter is not. So the claim in your bullet point holds only for sets of the form $A\times B$, right? (This is somewhat similar to the discussion here: Products of ultrafilters and products of measures.) $\endgroup$ – Martin Sleziak Sep 1 '18 at 0:36
  • $\begingroup$ I will add that this previous post of mine is related to the condition about limits from the first bullet point: Tensor product of ultrafilters corresponds to iterated limit. $\endgroup$ – Martin Sleziak Sep 1 '18 at 0:40
  • $\begingroup$ The construction of the measure from a filter seems to be faulty: unless $F$ is an ultrafilter, there should be a set $A \subseteq X$ such that neither $A \in F$ nor $X \setminus A \in F$, so we should have $p(A)=p(X\setminus A)=0$ but $p(A \cup (X \setminus A))=1$ so $p$ cannot be a measure at least not on the set-algebra of all the subsets of $X$. $\endgroup$ – Giorgio Mossa Apr 25 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.