# Are all linear transformations measurable?

Let $V$ and $W$ be topological vector spaces over $\mathbb{F}$ (with $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$), and let $T:V \to W$ be a linear transformation. It is well-known that $T$ is not necessarily continuous. But is $T$ necessarily measurable (with respect to the Borel structures of $V$ and $W$)? Does the answer change when we specifically consider $W=\mathbb{F}$?

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This fails for even finite-dimensional spaces. See math.stackexchange.com/questions/359183/… Or even better yet, see this: artsci.kyushu-u.ac.jp/~ssaito/eng/maths/Cauchy.pdf – Todd Trimble Aug 1 '14 at 12:06
@ToddTrimble: Thank you for your reply - but in the context of vector spaces, when I say "linear transformation", I mean a transformation that preserves both addition and scalar multiplication. – Julian Newman Aug 1 '14 at 15:03
Oh, of course you're right -- my apologies. – Todd Trimble Aug 1 '14 at 15:08

This question is answered by the well-known construction of a non-continuous linear form on an infinite dimensional Banach space using Hamel bases. Note also that there is a measurable graph theorem (L. Schwartz) which implies that all measurable linear maps, say between separable Banach spaces, are continuous. And there are models of set theory in which ALL linear mappings between suitable classes of spaces are continuous (Solovay and Garnir).

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Thank you for your reply. Sorry, I'm probably being a bit dim: how do you prove that the standard example of a discontinuous linear form on an infinite-dimensional Banach space is non-measurable? – Julian Newman Aug 1 '14 at 6:12
This follows by standard arguments from the fact that it is unbounded on any ball. – blackburne Aug 1 '14 at 6:35
I understand that the point of the example is that it is unbounded on any ball - but I do not know what the "standard arguments" are by which one can deduce non-measurability. (I presume you must be using completeness of the Banach space somehow, since on a countable-dimensional normed vector space the same construction gives a map that is unbounded on every ball and yet is measurable.) – Julian Newman Aug 1 '14 at 6:51
Baire's category theorem (cover the Banach space by the sets where $|f|\leq n$). By the way, it is not customary for questons of this sort to be answered in MO to the last detsil so this is my final word. – blackburne Aug 1 '14 at 7:05
Generalization (automatic continuity)... For a homomorphism between complete separable metric groups, if it has the property of Baire, then it is continuous. (In particular, Borel-Borel measurable maps have the property of Baire.) – Gerald Edgar Aug 1 '14 at 15:21