A counterexample on Pettis integrability #

There are several theories of integration for functions taking values in Banach spaces. Bochner integration, requiring approximation by simple functions, is the analogue of the one-dimensional theory. It is very well behaved, but only works for functions with second-countable range.

For functions f taking values in a larger Banach space B, one can define the Dunford integral as follows. Assume that, for all continuous linear functional φ, the function φ ∘ f is measurable (we say that f is weakly measurable, or scalarly measurable) and integrable. Then φ ↦ ∫ φ ∘ f is continuous (by the closed graph theorem), and therefore defines an element of the bidual B**. This is the Dunford integral of f.

This Dunford integral is not usable in practice as it does not belong to the right space. Let us say that a function is Pettis integrable if its Dunford integral belongs to the canonical image of B in B**. In this case, we define the Pettis integral as the Dunford integral inside B.

This integral is very general, but not really usable to do analysis. This file illustrates this, by giving an example of a function with nice properties but which is not Pettis-integrable. This function:

is defined from [0, 1] to a complete Banach space;
is weakly measurable;
has norm everywhere bounded by 1 (in particular, its norm is integrable);
and yet it is not Pettis-integrable with respect to Lebesgue measure.

This construction is due to Ralph S. Phillips, Integration in a convex linear topological space, in Example 10.8. It requires the continuum hypothesis. The example is the following.

Under the continuum hypothesis, one can find a subset of ℝ² which, along each vertical line, only misses a countable set of points, while it is countable along each horizontal line. This is due to Sierpinski, and formalized in sierpinski_pathological_family. (In fact, Sierpinski proves that the existence of such a set is equivalent to the continuum hypothesis).

Let B be the set of all bounded functions on ℝ (we are really talking about everywhere defined functions here). Define f : ℝ → B by taking f x to be the characteristic function of the vertical slice at position x of Sierpinski's set. It is our counterexample.

To show that it is weakly measurable, we should consider φ ∘ f where φ is an arbitrary continuous linear form on B. There is no reasonable classification of such linear forms (they can be very wild). But if one restricts such a linear form to characteristic functions, one gets a finitely additive signed "measure". Such a "measure" can be decomposed into a discrete part (supported on a countable set) and a continuous part (giving zero mass to countable sets). For all but countably many points, f x will not intersect the discrete support of φ thanks to the definition of the Sierpinski set. This implies that φ ∘ f is constant outside of a countable set, and equal to the total mass of the continuous part of φ there. In particular, it is measurable (and its integral is the total mass of the continuous part of φ).

Assume that f has a Pettis integral g. For all continuous linear form φ, then φ g should be the total mass of the continuous part of φ. Taking for φ the evaluation at the point x (which has no continuous part), one gets g x = 0. Take then for φ the Lebesgue integral on [0, 1] (or rather an arbitrary extension of Lebesgue integration to all bounded functions, thanks to Hahn-Banach). Then φ g should be the total mass of the continuous part of φ, which is 1. This contradicts the fact that g = 0, and concludes the proof that f has no Pettis integral.

Implementation notes #

The space of all bounded functions is defined as the space of all bounded continuous functions on a discrete copy of the original type, as mathlib only contains the space of all bounded continuous functions (which is the useful one).

A counterexample on Pettis integrability #

Implementation notes #

Extending the integral #

Additive measures on the space of all sets #

Relationship between continuous functionals and finitely additive measures. #

A set in ℝ² large along verticals, small along horizontals #

A counterexample for the Pettis integral #

A set in `ℝ²` large along verticals, small along horizontals #