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Topic:
From addition preservation to linearity
Replies:
3
Last Post:
Sep 24, 2012 1:14 PM




Re: From addition preservation to linearity
Posted:
Sep 24, 2012 12:15 PM


On Mon, 24 Sep 2012 01:41:01 0700 (PDT), Rupert <rupertmccallum@yahoo.com> wrote:
>On Sep 24, 1:14 am, José Carlos Santos <jcsan...@fc.up.pt> wrote: >> Hi all, >> >> Can someone please tell me where can I find a reference for the fact >> that any Lebesguemeasurable map from R^n into R^m which preserves sums >> is necessarily a linear map? >> >> Best regards, >> >> Jose Carlos Santos > >I don't know a reference, but one of the crucial points is that if A >is a subset of R^n of positive measure then A+A has nonempty interior. >David Ullrich showed me a proof of this a while back but he didn't >know a reference in the literature.
Hmm. It's unclear to me what the "this" refers to  I've posted the simple proof of that fact about A + A, but I don't recall ever saying anything about the OP's question. In fact it's not clear to me how that follows, let's see:
Lemma 0. If A subset R^n and m(A) > 0 then A + A has nonempty interior.
Pf Sketch: WLOG 0 < m(A) < infinity. Let h be the characteristic function of A, and let g be the convolution of A with itself. Then g >= 0, and
int g = (int h)^2 > 0.
Also g is continuous, so g > 0 on some nonempty open set B. But B subset A + A. QED.
Hmm. Now suppose f : R^n > R^m is measurable and
f(x+y) = f(x) + f(y).
There exists L so that if A = {x : f(x) < L} then m(L) > 0. Hence A + A has nonempty interior O.
But x in O implies that f(x) < 2L.
Fix x in O and let V = O  x = {y : x + y in O}. Then V is a neighborhood of the origin, and z in in V implies z = y  x with y in O, hence
f(z) < 2L + f(x) = c.
So f is bounded in O.
Now since f(nx) = n f(x) for any positive integer n, it follows that
f(z) < c/n for z in O/n,
and hence f is continuous at the origin. Since f is additive this shows that f is continuous.
And now, since f(qx) = q f(x) for any rational q and f is continuous, it follows that f(rx) = r f(x) for any real r, so f is linear. QED.
Huh. I didn't know I knew that, thanks.
> >I think this might be relevant > >@article{Kestelman51, > author="Hyman Kestelman", > title="Automorphisms of the field of complex >numbers", > journal="Proc. London Math. Soc.", > issue=2, > number=53, > pages="112", > year=1951}



