Editing Radon–Nikodym theorem (section)

== Examples ==
In the following examples, the set {{mvar|X}} is the real interval [0,1], and <math>\Sigma</math> is the [[Borel set|Borel sigma-algebra]] on {{mvar|X}}.
# <math>\mu</math> is the length measure on {{mvar|X}}. <math>\nu</math> assigns to each subset {{mvar|Y}} of {{mvar|X}}, twice the length of {{mvar|Y}}. Then, <math display="inline">\frac{d\nu}{d\mu} = 2</math>.
# <math>\mu</math> is the length measure on {{mvar|X}}. <math>\nu</math> assigns to each subset {{mvar|Y}} of {{mvar|X}}, the number of points from the set {0.1, …, 0.9} that are contained in {{mvar|Y}}. Then, <math>\nu</math> is not absolutely-continuous with respect to <math>\mu</math> since it assigns non-zero measure to zero-length points. Indeed, there is no derivative <math display="inline">\frac{d\nu}{d\mu}</math>: there is no finite function that, when integrated e.g. from <math>(0.1 - \varepsilon)</math> to <math>(0.1 + \varepsilon)</math>, gives <math>1</math> for all <math>\varepsilon > 0</math>.
# <math>\mu = \nu + \delta_0</math>, where <math>\nu</math> is the length measure on {{mvar|X}} and <math>\delta_0</math> is the [[Dirac measure]] on 0 (it assigns a measure of 1 to any set containing 0 and a measure of 0 to any other set). Then, <math>\nu</math> is absolutely continuous with respect to <math>\mu</math>, and <math display="inline">\frac{d\nu}{d\mu} = 1_{X\setminus \{0\}}</math> – the derivative is 0 at <math>x = 0</math> and 1 at <math>x > 0</math>.<ref>{{cite web |title=Calculating Radon Nikodym derivative |date=April 7, 2018 |work=[[Stack Exchange]] |url=https://math.stackexchange.com/q/588442/29780 }}</ref>