Take the 2-minute tour ×

Does anyone link me to a reference that proves the following theorem for Plessner:

Let $\mu$ be a bounded complex Borel measure on $\mathbb{R}$ $\mu$ is absolutely continuous to the Lebesgue measure $\Longleftrightarrow$ lim $\mu (E + t)$ = $\mu(E)$ for every Borel set E.

Thank you.

share|improve this question
A quick google search yields springerlink.com/content/2734828m01452hm3. Is this the kind of thing you were looking for? –  Peter Humphries Apr 28 '11 at 9:08
This link gives a distributional proof. Im asking for plessner's original. –  jessica Apr 30 '11 at 7:18
add comment

1 Answer

I know you said you were looking for Plessner's original, but you can give a short proof of this. Fix a compactly supported smooth function $\phi$ with integral one. Let $\phi_{t}(x)=t^{-1}\phi(x/t),$ and set $\nu_{t}=\mu*\phi_{t}.$ Note that $\nu_{t}$ is absolutely continuous with respect to Lebesgue with Radon-Nikodym derivative $\int\phi_{t}(x-y)\,d\mu(y).$ If $f$ is a $C_{0}(\Bbb R)$ function a straight-foward computation using Fubini's Theorem shows that $$\int fd\nu_{t}=\int\int f(y)\phi_{t}(y)d\tau_{x}\mu(y)dx$$ where $\tau_{x}\mu(E)=\mu(E-x).$ Because $\mu$ is continuous under translation the usual arguments show that $\nu_{t}$ converges to $\mu$ in norm, (here think of the total variation norm as the operator norm). Thus $\mu$ is a norm limit of $L^{1}(\Bbb R)$ functions and thus must be in $L^{1}(\Bbb R).$
(Note that the argument can be used to prove similar statements e.g. an $L^{\infty}$ function continuous under translations must agree almost everywhere with a uniformly continuous function).

share|improve this answer
Nice, this is much shorter than the proof I came up with. (P.s. don't forget rent). –  Kevin Ventullo May 2 '11 at 7:30
add comment

Your Answer


