Ngô Quốc Anh

August 17, 2009

The use of equivalence in measure

In mathematics, and specifically in measure theory, equivalence is a notion of two measures being “the same”. Two measures are equivalent if they have the same null sets.

Definition. Let (X, \Sigma) be a measurable space, and let \mu, \nu : \Sigma \to [0, +\infty] be two measures. Then \mu is said to be equivalent to \nu if

\mu (A) = 0 \iff \nu (A) = 0

for measurable sets A in \Sigma, i.e. the two measures have precisely the same null sets. Equivalence is often denoted \displaystyle{\mu \sim \nu} or \mu \approx \nu.

In terms of absolute continuity of measures, two measures are equivalent if and only if each is absolutely continuous with respect to the other:

\mu \sim \nu \iff \mu \ll \nu \ll \mu.

Equivalence of measures is an equivalence relation on the set of all measures \Sigma \to [0, +\infty].

Examples.

  1. Gaussian measure and Lebesgue measure on the real line are equivalent to one another.
  2. Lebesgue measure and Dirac measure on the real line are inequivalent.

Application. Let \mu be a finite measure on \mathbb R, and define

f\left( x \right) = \int\limits_{ - \infty }^{ + \infty } {\frac{{\ln \left| {x - t} \right|}} {{\sqrt {\left| {x - t} \right...

Show that f(x) is finite a.e. with respect to the Lebesgue measure on \mathbb R.

Proof. Let

g\left( x \right) = \left\{ \begin{gathered}   \frac{{\ln \left| x \right|}} {{\sqrt {\left| x \right|} }},x \ne 0, \hfill \\...

then g \in L^1(\mathbb R, d\nu) where

d\nu \left( x \right) = \frac{{dx}} {{1 + {x^2}}}

and

f\left( x \right) = \int_{ - \infty }^{ + \infty } {g\left( {x - t} \right)d\mu \left( t \right)} ,x \in \mathbb{R}.

Clearly

\int_{ - \infty }^{ + \infty } {\left| {f\left( x \right)} \right|d\mu \left( x \right)}  \leqslant \int_{ - \infty }^{ + \in...

and by Fubini’s Theorem

\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {\left| {g\left( {x - t} \right)} \right|d\mu \left( t...

then

\int_{ - \infty }^{ + \infty } {\left| {f\left( x \right)} \right|d\mu \left( x \right)}  \leqslant \int_{ - \infty }^{ + \in...

Since

\begin{gathered}   \int\limits_{ - \infty }^{ + \infty } {\left( {\int\limits_{ - \infty }^{ + \infty } {\left| {g\left( {x -...

Thus the following function

x \mapsto \int\limits_{ - \infty }^{ + \infty } {\left| {g\left( {x - t} \right)} \right|d\mu \left( t \right)}

finite a.e. with respect to the measure \nu. The conclusion follows from that the measure \nu and the Lebesgue measure are equivalent.

Source: http://en.wikipedia.org/wiki/Equivalence_(measure_theory)

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