logo

Radon-Nikodym Derivative 📂Measure Theory

Radon-Nikodym Derivative

Theorem 1

Assuming a measurable space (Ω,F)( \Omega , \mathcal{F} ) is given with measures μ\mu, ν\nu satisfying μ(Ω)=1\mu ( \Omega ) = 1 for all F∈FF \in \mathcal{F} and 0≤ν(F)≤μ(F)0 \le \nu (F) \le \mu (F), there exists a F\mathcal{F}-measurable function h:Ω→Rh : \Omega \to \mathbb{R} for all F∈FF \in \mathcal{F} that satisfies ν(F)=∫Fhdμ \nu (F) = \int_{F} h d \mu This hh, denoted as h:=dνdμ\displaystyle h := {{d \nu } \over {d \mu }}, is called the Radon-Nikodym derivative of ν\nu with respect to μ\mu.

  • A function ff being a F\mathcal{F}-measurable function means it satisfies f−1(B)∈Ff^{-1} (B) \in \mathcal{F} for all Borel sets B∈B(Ω)B \in \mathcal{B} ( \Omega ).

Explanation

Under the premise of the theorem that μ(Ω)=1\mu ( \Omega ) = 1, Ω\Omega can become a probability space (Ω,F,μ)( \Omega , \mathcal{F} , \mu ).

The naming of the Radon-Nikodym derivative is quite intuitive, even if putting the exact argument aside for a moment and dealing with it as it appears might unfold like so. Unlike differentiation in basic analysis, it’s a matter of fitting concepts to forms. ∫Fhdμ=∫Fdνdμdμ=∫Fdν=ν(F) \begin{align*} \int_{F} h d \mu =& \int_{F} {{d \nu } \over {d \mu }} d \mu \\ =& \int_{F} d \nu \\ =& \nu ( F) \end{align*}

The Radon-Nikodym theorem guarantees the unique existence of such Radon-Nikodym derivatives under a few more given conditions.

Proof

Part 1. hP=νμ\displaystyle h_{\mathcal{P}} = {{\nu} \over {\mu}}

Let’s define hP:Ω→Rh_{\mathcal{P}} : \Omega \to \mathbb{R} for partitions P:={A1,⋯ ,Ak} \mathcal{P} := \left\{ A_{1} , \cdots , A_{k} \right\} and ω∈Ai\omega \in A_{i} as follows. hP(ω)={ν(Ai)μ(Ai),μ(Ai)>00,otherwise h_{\mathcal{P}} ( \omega ) = \begin{cases} \displaystyle {{\nu (A_{i} ) } \over {\mu (A_{i} ) }} &, \mu ( A_{i} ) > 0 \\ 0 &, \text{otherwise} \end{cases} Given AiA_{i}, according to the definition, ω∈Ai\omega \in A_{i} can be treated as a constant function hP(ω)=cih_{\mathcal{P}} ( \omega ) = c_{i} regardless. Accordingly, ∫AihPdμ=∫Aiν(Ai)μ(Ai)dμ=ν(Ai)1μ(Ai)∫Aidμ=ν(Ai)1μ(Ai)μ(Ai)=ν(Ai) \begin{align*} \int_{A_{i}} h_{\mathcal{P}} d \mu =& \int_{A_{i}} {{\nu (A_{i} ) } \over {\mu (A_{i} ) }} d \mu \\ =& \nu (A_{i} ) {{1 } \over {\mu (A_{i} ) }} \int_{A_{i}} d \mu \\ =& \nu (A_{i} ) {{1 } \over {\mu (A_{i} ) }} \mu ( A_{i} ) \\ =& \nu (A_{i} ) \end{align*} Let’s prove the following corollaries.

  • Part 1-1.
    For all P\mathcal{P} of Ω\Omega and for all ω∈Ω\omega \in \Omega, since 0≤hP≤10 \le h_{\mathcal{P}} \le 1 for all F∈FF \in \mathcal{F} satisfies 0≤ν(F)≤μ(F)0 \le \nu (F) \le \mu (F), 0≤hP=ν(F)μ(F)≤1\displaystyle 0 \le h_{\mathcal{P}} = {{\nu (F)} \over {\mu (F)}} \le 1
  • Part 1-2.
    A=⨆j∈JAj  ⟹  ν(A)=∫AhPdμA = \bigsqcup_{j \in J} A_{j} \implies \nu (A) = \int_{A} h_{\mathcal{P}} d \mu ⨆\bigsqcup is a disjoint union and J⊂{1,⋯ ,k}J \subset \left\{ 1 , \cdots, k \right\} is the set of indices. By the properties of measure, ν(A)=∑j∈Jν(Aj)=∑j∈J∫AjhPdμ=∫AhPdμ \begin{align*} \nu (A) =& \sum_{j \in J} \nu ( A_{j} ) \\ =& \sum_{j \in J} \int_{A_{j}} h_{\mathcal{P}} d \mu \\ =& \int_{A} h_{\mathcal{P}} d \mu \end{align*} Meanwhile, since Ω∈F\Omega \in \mathcal{F} by the definition of the sigma field, naturally ν(Ω)=∫ΩhPdμ\displaystyle \nu ( \Omega ) = \int_{ \Omega } h_{\mathcal{P}} d \mu holds true.
  • Part 1-3.
    • Part 1-3-1. For all A∈P1A \in \mathcal{P}_{1}, ∫Ah1dμ=∫Ah2dμ\displaystyle \int_{A} h_{1} d \mu = \int_{A} h_{2} d \mu holds true.
      Let’s say P2\mathcal{P}_{2} is a refinement of P1\mathcal{P}_{1}, and for convenience, denote it as hn:=hPnh_{n} := h_{\mathcal{P}_{n}}. By the definition of refinement, there exist Bj∈P2B_{j} \in \mathcal{P}_{2} that satisfy A=⨆j∈JBj\displaystyle A = \bigsqcup_{j \in J} B_{j} for all A∈P1A \in \mathcal{P}_{1}. Therefore, ∫Ah1dμ=ν(A)=∑j∈Jν(Bj)=∑j∈J∫Bjh2dμ=∫Ah2dμ \begin{align*} \int_{A} h_{1} d \mu =& \nu (A) \\ =& \sum_{j \in J} \nu ( B_{j} ) \\ =& \sum_{j \in J} \int_{B_{j}} h_{2} d \mu \\ =& \int_{A} h_{2} d \mu \end{align*}
    • Part 1-3-2. For all A∈P1A \in \mathcal{P}_{1}, ∫Ah1h2dμ=∫Ah12dμ\displaystyle \int_{A} h_{1} h_{2} d \mu = \int_{A} h_{1}^{2} d \mu holds true. ∫Ah1h2dμ=ν(A)μ(A)∫Ah2dμ=ν(A)μ(A)ν(A)=∫A[ν(A)μ(A)]2dμ=∫Ah12dμ\begin{align*} \int_{A} h_{1} h_{2} d \mu =& {{\nu (A ) } \over {\mu (A ) }} \int_{A} h_{2} d \mu \\ =& {{\nu (A ) } \over {\mu (A ) }} \nu (A ) \\ =& \int_{A} \left[ {{\nu (A ) } \over {\mu (A ) }} \right]^2 d \mu \\ =& \int_{A} h_{1}^{2} d \mu \end{align*}
  • Part 1-4.
    • Part 1-4-1. ∫A(h2−h1)2dμ=∫A[h22−h12]dμ\displaystyle \int_{A} ( h_{2} - h_{1} )^2 d \mu = \int_{A} \left[ h_{2}^{2} - h_{1}^{2} \right] d \mu
      According to Part 1-3-2, for all A∈P1A \in \mathcal{P}_{1}, ∫Ah1(h2−h1)dμ=0\displaystyle \int_{A} h_{1} ( h_{2} - h_{1} ) d \mu = 0 holds true, and according to Part 1-2, ∫Ωh1(h2−h1)dμ=0\displaystyle \int_{ \Omega } h_{1} ( h_{2} - h_{1} ) d \mu = 0 also stands. Thus, ∫A(h2−h1)2dμ=∫A(h22−2h2h1+h12)dμ=∫A[h22−2h1(h2−h1)−h12]dμ=∫A[h22−h12]dμ \begin{align*} \int_{A} ( h_{2} - h_{1} )^2 d \mu =& \int_{A} \left( h_{2}^{2} - 2 h_{2} h_{1} + h_{1}^{2} \right) d \mu \\ =& \int_{A} \left[ h_{2}^{2} - 2 h_{1} (h_{2} - h_{1}) - h_{1}^{2} \right] d \mu \\ =& \int_{A} \left[ h_{2}^{2} - h_{1}^{2} \right] d \mu \end{align*}
    • Part 1-4-2. ∫Ωh12dμ≤∫Ωh22dμ\displaystyle \int_{\Omega} h_{1}^{2} d \mu \le \int_{\Omega} h_{2}^{2} d \mu
      ∫Ωh22dμ=∫Ωh12dμ+∫Ω(h2−h1)2dμ≥∫Ωh12dμ\begin{align*} \int_{\Omega} h_{2}^{2} d \mu =& \int_{\Omega} h_{1}^{2} d \mu + \int_{\Omega} (h_{2} - h_{1})^{2} d \mu \\ \ge& \int_{\Omega} h_{1}^{2} d \mu \end{align*}

Part 2. h:=lim⁡n→∞hQn\displaystyle h := \lim_{n \to \infty} h_{\mathcal{Q}_{n}}

In Part 1-4-2, it was confirmed that ∫ΩhP2dμ≤∫ΩhP′2dμ\displaystyle \int_{\Omega} h_{\mathcal{P}}^{2} d \mu \le \int_{\Omega} h_{\mathcal{P} ' }^{2} d \mu holds for the refinement P′\mathcal{P} ' of P\mathcal{P}. Moreover, since 0≤hP≤10 \le h_{\mathcal{P}} \le 1 in Part 1-1 and the assumption was μ(Ω)=1\mu ( \Omega ) = 1, c:=sup⁡∫ΩhP2dμc := \sup \int_{\Omega} h_{\mathcal{P}}^{2} d \mu exists between 00 and 11. [ NOTE: In fact, even if it’s μ(Ω)≠1\mu ( \Omega ) \ne 1, it can be substituted as μ′:=μ/μ(Ω)\mu ' := \mu / \mu ( \Omega). ] Now consider a partition Qn\mathcal{Q}_{n} that becomes a refinement of both given n∈Nn \in \mathbb{N} against ∫ΩhPn2dμ>c−14n\displaystyle \int_{\Omega} h_{\mathcal{P}_{n}}^{2} d \mu > c - {{1} \over {4^{n}}}. Naturally, Qn+1\mathcal{Q}_{n+1} is a refinement of Qn\mathcal{Q}_{n} and satisfies the following inequality. c−14n≤∫ΩhPn2dμ≤∫ΩhQn2dμ≤∫ΩhQn+12dμ≤c c - {{1} \over {4^{n}}} \le \int_{\Omega} h_{\mathcal{P}_{n}}^{2} d \mu \le \int_{\Omega} h_{\mathcal{Q}_{n}}^{2} d \mu \le \int_{\Omega} h_{\mathcal{Q}_{n+1}}^{2} d \mu \le c According to Part 1-4-1, since the square 2^2 can enter into the parenthesis, ∫Ω(hQn+1−hQn)2dμ=∫Ω(hQn+12−hQn2)dμ<14n \int_{\Omega} \left( h_{\mathcal{Q}_{n+1}} - h_{\mathcal{Q}_{n}} \right)^{2} d \mu = \int_{\Omega} \left( h_{\mathcal{Q}_{n+1}}^{2} - h_{\mathcal{Q}_{n}}^{2} \right) d \mu < {{1} \over {4^{n}}}

Cauchy-Schwarz Inequality: If f,g∈L2(E)f,g \in \mathcal{L}^{2} (E), then fg∈L1(E)fg \in L^{1}(E), and ∥∫Efg‾dm∥≤∥fg∥1≤∥f∥2∥g∥2 \left\| \int_{E} f \overline{g} dm \right\| \le \left\| f g \right\|_{1} \le \left\| f \right\|_{2} \left\| g \right\|_{2}

Setting f=∣hQn+1−hQn∣f = | h_{\mathcal{Q}_{n+1}} - h_{\mathcal{Q}_{n}} |, g=1g = 1 in the Cauchy-Schwarz inequality, for all n∈Nn \in \mathbb{N}, ∫Ω∣hQn+1−hQn∣dμ≤∫Ω(hQn+1−hQn)2dμ∫Ω1dμ=∫Ω(hQn+12−hQn2)dμμ(Ω)<12n⋅1 \begin{align*} \int_{\Omega} \left| h_{\mathcal{Q}_{n+1}} - h_{\mathcal{Q}_{n}} \right| d \mu \le & \sqrt{ \int_{\Omega} \left( h_{\mathcal{Q}_{n+1}} - h_{\mathcal{Q}_{n}} \right)^{2} d \mu } \sqrt{ \int_{\Omega} 1 d \mu } \\ =& \sqrt{\int_{\Omega} \left( h_{\mathcal{Q}_{n+1}}^{2} - h_{\mathcal{Q}_{n}}^{2} \right) d \mu} \sqrt{ \mu ( \Omega ) } \\ <& {{1} \over {2^{n}}} \cdot 1 \end{align*}

Levi’s Theorem: If ∑k=1∞∫∣fk∣dm<∞\displaystyle \sum_{k=1}^{\infty} \int |f_{k}| dm < \infty, then ∑k=1∞fk(x)\displaystyle \sum_{k=1}^{\infty} f_{k} (x) converges almost everywhere and ∫∑k=1∞fkdm=∑k=1∞∫fkdm \int \sum_{k=1}^{\infty} f_{k} dm = \sum_{k=1}^{\infty} \int f_{k} dm

Since ∑n∈N∫Ω∣hQn+1−hQn∣dμ<∞\displaystyle \sum_{n \in \mathbb{N}} \int_{\Omega} \left| h_{\mathcal{Q}_{n+1}} - h_{\mathcal{Q}_{n}} \right| d \mu < \infty, by Levi’s theorem, ∑n∈N(hQn+1−hQn)=lim⁡n→∞hQn−hQ1 \sum_{n \in \mathbb{N}} \left( h_{\mathcal{Q}_{n+1}} - h_{\mathcal{Q}_{n}} \right) = \lim_{n \to \infty} h_{\mathcal{Q}_{n}} - h_{\mathcal{Q}_{1}} converges almost everywhere with respect to μ\mu. Now let’s define hh as follows. h:=hQ1+∑n∈N(hQn+1−hQn)=lim⁡n→∞hQn h := h_{\mathcal{Q}_{1}} + \sum_{n \in \mathbb{N}} \left( h_{\mathcal{Q}_{n+1}} - h_{\mathcal{Q}_{n}} \right) = \lim_{n \to \infty} h_{\mathcal{Q}_{n}}

Part 3. ν(F)=∫Fhdμ\displaystyle \nu (F) = \int_{F} h d \mu

By the definition of hh, 0≤h≤10 \le h \le 1 is F\mathcal{F}-measurable. Now it suffices to show that ν(F)=∫Fhdμ\displaystyle \nu (F) = \int_{F} h d \mu for all F∈FF \in \mathcal{F}. Fix one F∈FF \in \mathcal{F} and define Rn\mathcal{R}_{n} as a common refinement partition of Qn\mathcal{Q}_{n} and {F,Fc}\left\{ F , F^{c} \right\}. Then, according to A=⨆j∈JAj  ⟹  ν(A)=∫AhPdμ\displaystyle A = \bigsqcup_{j \in J} A_{j} \implies \nu (A) = \int_{A} h_{\mathcal{P}} d \mu shown in Part 1-2, for all n∈Nn \in \mathbb{N}, ν(F)=∫FhRndμ=∫F(hRn−hQn)dμ+∫FhQndμ \begin{align*} \nu (F) =& \int_{F} h_{ \mathcal{R}_{n} } d \mu \\ =& \int_{F} ( h_{ \mathcal{R}_{n} } - h_{ \mathcal{Q}_{n} } )d \mu + \int_{F} h_{ \mathcal{Q}_{n} } d \mu \end{align*} is satisfied. Meanwhile, using the Cauchy-Schwarz inequality for Rn\mathcal{R}_{n} as in Part 2, we can derive ∣∫ΩhRn−hQndμ∣<12n\displaystyle \left| \int_{\Omega} h_{\mathcal{R}_{n}} - h_{\mathcal{Q}_{n}} d \mu \right| < {{1} \over {2^{n}}}, thus ν(F)=0+lim⁡n→∞∫FhQndμ \nu (F) =0 + \lim_{n \to \infty} \int_{F} h_{ \mathcal{Q}_{n} } d \mu

Dominated Convergence Theorem: For measurable sets E∈ME \in \mathcal{M} and g∈L1(E)g \in \mathcal{L}^{1} (E), suppose a sequence of measurable functions {fn}\left\{ f_{n} \right\} satisfies ∣fn∣≤g|f_{n}| \le g almost everywhere on EE. If almost everywhere on EE, f=lim⁡n→∞fn\displaystyle f = \lim_{n \to \infty} f_{n} holds, then f∈L1(E)f \in \mathcal{L}^{1}(E) and lim⁡n→∞∫Efn(x)dm=∫Efdm \lim_{ n \to \infty} \int_{E} f_{n} (x) dm = \int_{E} f dm

Since 0≤hQn≤10 \le h_{\mathcal{Q}_{n}} \le 1 for all n∈Nn \in \mathbb{N} and hh was defined as such in Part 2, by the Dominated Convergence Theorem, ν(F)=lim⁡n→∞∫FhQndμ=∫Fhdμ \begin{align*} \nu (F) =& \lim_{n \to \infty} \int_{F} h_{ \mathcal{Q}_{n} } d \mu \\ =& \int_{F} h d \mu \end{align*}

■

Meanwhile, from Part 1~2 of the proof above, we derive the following corollary.

Corollary

If for all n∈Nn \in \mathbb{N}, Qn+1\mathcal{Q}_{n+1} is a refinement of Qn\mathcal{Q}_{n}, lim⁡n→∞hQn=lim⁡n→∞νμ=dνdμ \lim_{n \to \infty} h_{\mathcal{Q}_{n}} = \lim_{n \to \infty} {{\nu} \over {\mu}} = {{d \nu } \over {d \mu }}

  1. Folland. (1999). Real Analysis: Modern Techniques and Their Applications(2nd Edition): p91. ↩︎