A nonnegative measurable function \(f\) turns one measure into another: \(\nu(E)=\int_Ef\,d\mu\) is a measure, by Proposition 4.6, and \(f\) is its density. Mass distributions with a density per unit length, charges with a density per unit volume, and weighted counts are all of this form. The question of this chapter is the converse. Given two measures \(\mu\) and \(\nu\) on the same measurable space, is \(\nu\) obtained from \(\mu\) by a density?
One obstruction is plain. If \(\mu(E)=0\) then \(\int_Ef\,d\mu=0\) whatever \(f\) is, so a measure with a density must vanish on every \(\mu\) -null set. A point mass on the line has no density with respect to length, because it charges a set of length zero. The Radon–Nikodym theorem says that, for \(\sigma\) -finite measures, this is the only obstruction.
The proof has to compare two measures set by set, and the natural way to compare \(\nu\) with a multiple of \(\mu\) is to look at the difference \(\nu-c\mu\) and ask where it is positive. A difference of measures is not a measure; it can take negative values. So the chapter begins with set functions of both signs, and its first theorem says that the space always splits into a part where such a function is positive and a part where it is negative.
9.1 Signed measures and the Hahn decomposition
Definition 9.1 (signed measure). A signed measure on a measurable space \((X,\mathcal A)\) is a function \(\nu:\mathcal A\to[-\infty,\infty]\) that takes at most one of the values \(+\infty\) and \(-\infty\) , satisfies \(\nu(\emptyset)=0\) , and is countably additive: \(\nu(\bigcup_nE_n)=\sum_n\nu(E_n)\) for pairwise disjoint \(E_n\in\mathcal A\) .
To avoid confusion, a measure in the earlier sense will be called a positive measure in this chapter. Two kinds of example cover everything, as it turns out: the difference \(\nu_1-\nu_2\) of two positive measures at least one of which is finite; and \(\nu(E)=\int_Ef\,d\mu\) , where \(\mu\) is a positive measure and \(f\) is a measurable function with \(\int f^-\,d\mu<\infty\) or \(\int f^+\,d\mu<\infty\) .
A signed measure is not monotone: a subset can have larger measure than the set containing it. What survives is continuity along increasing sequences, with the proof of Theorem 1.10(a), and the fact that a subset of a set of finite measure has finite measure, because \(\nu(F)=\nu(E)+\nu(F\setminus E)\) and at most one infinite value is available.
Definition 9.2 (positive, negative, and null sets). Let \(\nu\) be a signed measure. A set \(P\in\mathcal A\) is positive for \(\nu\) if \(\nu(E)\ge0\) for every measurable \(E\subseteq P\) ; negative if \(\nu(E)\le0\) for every measurable \(E\subseteq P\) ; and null if \(\nu(E)=0\) for every measurable \(E\subseteq P\) .
A set of positive measure need not be a positive set: it may contain subsets of negative measure. Measurable subsets of positive sets are positive, by definition, and so are countable unions: if \(P_1,P_2,\dots\) are positive and \(E\subseteq\bigcup_nP_n\) , then disjointification writes \(E\) as a disjoint union of sets \(E_n\subseteq P_n\) , each of nonnegative measure.
Lemma 9.3 (a positive set inside a set of positive measure). Let \(\nu\) be a signed measure that does not take the value \(+\infty\) , and let \(E\in\mathcal A\) with \(\nu(E)>0\) . Then \(E\) contains a positive set \(P\) with \(\nu(P)>0\) .
Proof. If \(E\) is positive, take \(P=E\) . Otherwise remove subsets of negative measure, greedily. Let \(n_1\) be the least positive integer for which some measurable \(E_1\subseteq E\) has \(\nu(E_1)<-\tfrac1{n_1}\) , and choose such an \(E_1\) . If \(E\setminus E_1\) is positive, stop. Otherwise let \(n_2\) be the least positive integer for which some \(E_2\subseteq E\setminus E_1\) has \(\nu(E_2)<-\tfrac1{n_2}\) , and continue. If the process stops after finitely many steps, the remaining set is positive, and its measure is \(\nu(E)\) minus a sum of negative numbers, hence positive.
If it never stops, put \(P=E\setminus\bigcup_kE_k\) . By countable additivity
Here \(\nu(E)\) is finite and \(\nu(P)<\infty\) , so the series converges; its terms are below \(-\tfrac1{n_k}\) , so \(\sum_k\tfrac1{n_k}<\infty\) and \(n_k\to\infty\) . Also \(\nu(P)\ge\nu(E)>0\) . To see that \(P\) is positive, suppose some measurable \(A\subseteq P\) had \(\nu(A)<0\) . Choose \(k\) with \(n_k\ge2\) and \(\tfrac1{n_k-1}<-\nu(A)\) . Then \(A\subseteq E\setminus(E_1\cup\dots\cup E_{k-1})\) and \(\nu(A)<-\tfrac1{n_k-1}\) , contradicting the minimality of \(n_k\) .
Theorem 9.4 (Hahn decomposition). Let \(\nu\) be a signed measure on \((X,\mathcal A)\) . There are a positive set \(P\) and a negative set \(N\) with \(P\cup N=X\) and \(P\cap N=\emptyset\) . If \(P',N'\) is another such pair, then \(P\triangle P'\) is a null set for \(\nu\) .
Proof. Replacing \(\nu\) by \(-\nu\) if necessary, assume \(\nu\) does not take the value \(+\infty\) . Let \(m=\sup\{\nu(Q):Q\ \text{positive}\}\) , which is at least \(0\) because \(\emptyset\) is positive. Choose positive sets \(Q_j\) with \(\nu(Q_j)\to m\) and let \(P=\bigcup_jQ_j\) . Then \(P\) is positive, so \(\nu(P)=\nu(Q_j)+\nu(P\setminus Q_j)\ge\nu(Q_j)\) for each \(j\) , whence \(\nu(P)=m\) , and \(m<\infty\) .
Let \(N=X\setminus P\) . If \(N\) were not negative, it would contain a set \(E\) with \(\nu(E)>0\) , and by Lemma 9.3 a positive set \(P_0\subseteq E\) with \(\nu(P_0)>0\) . Then \(P\cup P_0\) would be a positive set with \(\nu(P\cup P_0)=m+\nu(P_0)>m\) , contradicting the definition of \(m\) .
For uniqueness, \(P\setminus P'\subseteq P\cap N'\) is both positive and negative, hence null; likewise \(P'\setminus P\) .
Theorem 9.5 (Jordan decomposition). Let \(\nu\) be a signed measure with Hahn decomposition \(X=P\cup N\) . The set functions
are positive measures, at least one of them finite, with \(\nu=\nu^+-\nu^-\) . They do not depend on the choice of Hahn decomposition.
Proof. Both are countably additive because \(\nu\) is, nonnegative because \(P\) is positive and \(N\) negative, and \(\nu(E)=\nu(E\cap P)+\nu(E\cap N)\) . Since \(\nu\) omits one infinite value, one of \(\nu^{\pm}\) is finite. If \(P'\) is the positive set of another decomposition, then \(\nu(E\cap P)\) and \(\nu(E\cap P')\) differ by the measures of two subsets of the null set \(P\triangle P'\) .
The measures \(\nu^+\) and \(\nu^-\) are the positive and negative variations of \(\nu\) , and \(\lvert\nu\rvert=\nu^++\nu^-\) is its total variation. They live on disjoint sets, a relation that has a name.
Definition 9.6 (mutually singular measures). Two positive measures \(\mu\) and \(\nu\) on \((X,\mathcal A)\) are mutually singular, written \(\mu\perp\nu\) , if there is a set \(A\in\mathcal A\) with \(\mu(A)=0\) and \(\nu(X\setminus A)=0\) .
So \(\nu^+\perp\nu^-\) , and the Jordan decomposition is the unique way of writing \(\nu\) as a difference of two mutually singular positive measures (Exercise 9.2 has the argument).
Example 9.1: A signed measure on five points
Let \(X=\{1,2,3,4,5\}\) with all subsets measurable and let \(\nu\) have the point values \(3,-1,2,-4,0\) . A set is positive exactly when it contains neither \(2\) nor \(4\) , and negative exactly when it contains neither \(1\) nor \(3\) . So there are two Hahn decompositions,
differing by the null set \(\{5\}\) . Either gives the same variations: \(\nu^+\) has point values \(3,0,2,0,0\) and \(\nu^-\) has \(0,1,0,4,0\) . Hence
The whole space has measure zero and is very far from null. Among the \(32\) subsets, the largest value of \(\nu\) is \(5=\nu^+(X)\) , attained exactly at the two positive sets \(\{1,3\}\) and \(\{1,3,5\}\) , and the smallest is \(-5\) . The set \(\{1,2\}\) has positive measure \(2\) but is not a positive set, since it contains \(\{2\}\) .
Example 9.2: A signed measure given by a density
On \([0,2\pi]\) let \(\nu(E)=\int_E\sin x\,dx\) . A set on which \(\sin\ge0\) is positive, so \(P=[0,\pi]\) and \(N=(\pi,2\pi]\) form a Hahn decomposition, and
In general, if \(\nu(E)=\int_Ef\,d\mu\) , then \(\{f\ge0\}\) and \(\{f<0\}\) form a Hahn decomposition, \(\nu^{\pm}\) have the densities \(f^{\pm}\) , and \(\lvert\nu\rvert\) has the density \(\lvert f\rvert\) . The Jordan decomposition of a measure is the abstract form of splitting a function into its positive and negative parts, which is why the notation agrees.
9.2 Absolute continuity
Definition 9.7 (absolute continuity). Let \(\mu\) be a positive measure and \(\nu\) a positive or signed measure on \((X,\mathcal A)\) . Then \(\nu\) is absolutely continuous with respect to \(\mu\) , written \(\nu\ll\mu\) , if \(\nu(E)=0\) for every \(E\in\mathcal A\) with \(\mu(E)=0\) .
Every measure with a density is absolutely continuous. Absolute continuity and mutual singularity are opposite extremes: if \(\nu\ll\mu\) and \(\nu\perp\mu\) , then \(\nu=0\) , since \(\nu\) is carried by a \(\mu\) -null set, on which it must vanish. The relation is not symmetric. On the line, the measure with density \(\mathbf 1_{[0,1]}\) is absolutely continuous with respect to \(\lambda\) , but \(\lambda\) is not absolutely continuous with respect to it.
The name comes from the following quantitative form, which for \(\nu(E)=\int_E\lvert f\rvert\,d\mu\) is Proposition 4.10.
Proposition 9.8 (the epsilon–delta form). Let \(\mu\) be a positive measure and \(\nu\) a finite positive measure on \((X,\mathcal A)\) . Then \(\nu\ll\mu\) if and only if for every \(\varepsilon>0\) there is a \(\delta>0\) such that \(\mu(E)<\delta\) implies \(\nu(E)<\varepsilon\) .
Proof. If the condition holds and \(\mu(E)=0\) , then \(\nu(E)<\varepsilon\) for every \(\varepsilon\) , so \(\nu(E)=0\) . Conversely, suppose the condition fails: there are \(\varepsilon>0\) and sets \(E_n\) with \(\mu(E_n)<2^{-n}\) and \(\nu(E_n)\ge\varepsilon\) . Let \(F=\limsup_nE_n=\bigcap_n\bigcup_{k\ge n}E_k\) . By Proposition 1.12, \(\mu(F)=0\) . On the other hand \(\nu\bigl(\bigcup_{k\ge n}E_k\bigr)\ge\nu(E_n)\ge\varepsilon\) for every \(n\) , and since \(\nu\) is finite, continuity from above gives \(\nu(F)\ge\varepsilon\) . So \(\nu\) is not absolutely continuous.
Example 9.3: The quantitative form, and its failure for an infinite measure
On \((0,1)\) let \(\nu(E)=\int_Ex^{-1/2}\,dx\) , a finite measure with \(\nu((0,1))=2\) and \(\nu\ll\lambda\) . Because the density decreases, among all sets of a given Lebesgue measure \(\delta\) the interval \((0,\delta)\) has the largest \(\nu\) -measure, namely
So \(\lambda(E)<\delta\) implies \(\nu(E)<2\sqrt\delta\) , and to guarantee \(\nu(E)<\varepsilon\) it suffices to take \(\delta=\varepsilon^2/4\) . For \(\varepsilon=0.1\) this is \(\delta=0.0025\) .
Now let \(\rho(E)=\int_Ex^{-1}\,dx\) on \((0,1)\) . Again \(\rho\ll\lambda\) , since a set of Lebesgue measure zero has integral zero. But \(\rho((0,\delta))=\infty\) for every \(\delta>0\) : no \(\delta\) works for any \(\varepsilon\) . The finiteness of \(\nu\) in Proposition 9.8 cannot be dropped, and the proof shows why, in its appeal to continuity from above.
9.3 The Radon–Nikodym theorem
The proof rests on one comparison lemma, which is where the Hahn decomposition is used.
Lemma 9.9 (comparison of two finite measures). Let \(\mu\) and \(\nu\) be finite positive measures on \((X,\mathcal A)\) . Either \(\nu\perp\mu\) , or there are \(\varepsilon>0\) and a set \(E\in\mathcal A\) with \(\mu(E)>0\) such that \(\nu(A)\ge\varepsilon\mu(A)\) for every measurable \(A\subseteq E\) .
Proof. For each \(n\in\mathbb N\) let \(X=P_n\cup N_n\) be a Hahn decomposition for the signed measure \(\nu-\tfrac1n\mu\) . Put \(P=\bigcup_nP_n\) and \(N=\bigcap_nN_n=X\setminus P\) . The set \(N\) is negative for every \(\nu-\tfrac1n\mu\) , so \(0\le\nu(N)\le\tfrac1n\mu(N)\) for all \(n\) , and therefore \(\nu(N)=0\) . If \(\mu(P)=0\) , then \(\nu\perp\mu\) . Otherwise \(\mu(P_n)>0\) for some \(n\) , and \(P_n\) is a positive set for \(\nu-\tfrac1n\mu\) : take \(E=P_n\) and \(\varepsilon=\tfrac1n\) .
Theorem 9.10 (Lebesgue decomposition and Radon–Nikodym). Let \(\mu\) and \(\nu\) be \(\sigma\) -finite positive measures on \((X,\mathcal A)\) . Then there are unique positive measures \(\nu_a\) and \(\nu_s\) with
and there is a measurable function \(f:X\to[0,\infty)\) , unique up to \(\mu\) -null sets, such that \(\nu_a(E)=\int_Ef\,d\mu\) for all \(E\in\mathcal A\) . In particular, if \(\nu\ll\mu\) then \(\nu(E)=\int_Ef\,d\mu\) for all \(E\in\mathcal A\) .
Proof. Existence for finite measures. Let \(\mathcal F\) be the family of measurable \(g:X\to[0,\infty]\) such that \(\int_Eg\,d\mu\le\nu(E)\) for every \(E\in\mathcal A\) . It contains \(0\) . It is closed under maxima: if \(g,h\in\mathcal F\) and \(A=\{g>h\}\) , then for any \(E\) ,
Let \(a=\sup\{\int g\,d\mu:g\in\mathcal F\}\le\nu(X)<\infty\) , choose \(g_n\in\mathcal F\) with \(\int g_n\,d\mu\to a\) , and put \(f_n=\max(g_1,\dots,g_n)\) . Then \(f_n\in\mathcal F\) , the sequence increases, and \(\int f_n\,d\mu\to a\) . Let \(f=\sup_nf_n\) . By monotone convergence, \(\int_Ef\,d\mu=\lim_n\int_Ef_n\,d\mu\le\nu(E)\) for every \(E\) , so \(f\in\mathcal F\) , and \(\int f\,d\mu=a\) . Since \(a<\infty\) , \(f\) is finite \(\mu\) -a.e., and we may change it on a null set to be finite everywhere.
Define \(\nu_a(E)=\int_Ef\,d\mu\) and \(\nu_s(E)=\nu(E)-\nu_a(E)\) . Then \(\nu_a\ll\mu\) , and \(\nu_s\) is a finite positive measure because \(f\in\mathcal F\) . We claim \(\nu_s\perp\mu\) . If not, Lemma 9.9 gives \(\varepsilon>0\) and \(E_0\) with \(\mu(E_0)>0\) and \(\nu_s(A)\ge\varepsilon\mu(A)\) for all \(A\subseteq E_0\) . Then \(f+\varepsilon\mathbf 1_{E_0}\in\mathcal F\) , because for every \(E\)
But \(\int(f+\varepsilon\mathbf 1_{E_0})\,d\mu=a+\varepsilon\mu(E_0)>a\) , contradicting the definition of \(a\) .
Existence for \(\sigma\) -finite measures. Write \(X\) as a disjoint union of sets \(X_j\in\mathcal A\) with \(\mu(X_j)<\infty\) and \(\nu(X_j)<\infty\) , by intersecting the sets of a decomposition for \(\mu\) with those of one for \(\nu\) . Apply the finite case to the restrictions of \(\mu\) and \(\nu\) to each \(X_j\) , obtaining \(f_j\) and a \(\mu\) -null set \(B_j\subseteq X_j\) carrying the singular part there. Put \(f=\sum_jf_j\mathbf 1_{X_j}\) and \(B=\bigcup_jB_j\) , and define \(\nu_a(E)=\int_Ef\,d\mu\) and \(\nu_s(E)=\nu(E\cap B)\) . Then \(\mu(B)=0\) . On each \(X_j\) the absolutely continuous part of the restriction vanishes on the \(\mu\) -null set \(B_j\) and the singular part vanishes off \(B_j\) , so \(\nu(E\cap B_j)\) is the singular part of \(E\cap X_j\) and \(\nu(E\cap X_j\setminus B_j)=\int_{E\cap X_j}f_j\,d\mu\) . Summing over \(j\) gives \(\nu=\nu_a+\nu_s\) , with \(\nu_s\) carried by \(B\) .
Uniqueness of the decomposition. Let \(\nu=\nu_a+\nu_s=\nu_a'+\nu_s'\) be two decompositions, with \(\nu_s\) carried by the \(\mu\) -null set \(B\) and \(\nu_s'\) by the \(\mu\) -null set \(B'\) . Then \(C=B\cup B'\) is \(\mu\) -null, so \(\nu_a(C)=\nu_a'(C)=0\) , while \(\nu_s\) and \(\nu_s'\) vanish outside \(C\) . Hence for every \(E\) ,
Uniqueness of the density. Suppose \(\int_Ef\,d\mu=\int_Eg\,d\mu\) for all \(E\) , with \(f,g\) finite and nonnegative. For integers \(j,k,m\) let \(E=X_j\cap\{g\le m\}\cap\{f>g+\tfrac1k\}\) . Both integrals over \(E\) are finite, so \(0=\int_E(f-g)\,d\mu\ge\tfrac1k\mu(E)\) and \(\mu(E)=0\) . The union of these sets over \(j,k,m\) is \(\{f>g\}\) . So \(f\le g\) a.e., and by symmetry \(f=g\) a.e.
Definition 9.11 (Radon–Nikodym derivative). When \(\nu\ll\mu\) , the function \(f\) of Theorem 9.10 is the Radon–Nikodym derivative of \(\nu\) with respect to \(\mu\) , written \(\dfrac{d\nu}{d\mu}\) . It is defined up to \(\mu\) -null sets.
For a signed measure \(\nu\ll\mu\) with \(\lvert\nu\rvert\) \(\sigma\) -finite, apply the theorem to \(\nu^+\) and \(\nu^-\) and subtract. The decomposition \(\nu=\nu_a+\nu_s\) is the Lebesgue decomposition of \(\nu\) with respect to \(\mu\) .
The proof is a maximization. The family \(\mathcal F\) consists of the densities that never overcharge any set, and the Radon–Nikodym derivative is its largest member. Lemma 9.9 says that if anything absolutely continuous were left over after subtracting \(f\) , a little more density could be added, and so \(f\) would not have been the largest.
Common mistake. Writing \(\dfrac{d\nu}{d\mu}\) before checking the two hypotheses. Absolute continuity must be verified on null sets, and both measures must be \(\sigma\) -finite. The symbol is not a quotient of the numbers \(\nu(E)\) and \(\mu(E)\) for any particular set, and it has no meaning when \(\nu\) charges a \(\mu\) -null set.
Example 9.4: A density and a singular part on four points
Let \(X=\{1,2,3,4\}\) with all subsets measurable, and let \(\mu\) and \(\nu\) have the point masses
The only nonempty \(\mu\) -null set is \(\{4\}\) , and \(\nu(\{4\})=\tfrac38\ne0\) , so \(\nu\) is not absolutely continuous with respect to \(\mu\) . The Lebesgue decomposition puts the mass at \(4\) into the singular part:
On the points where \(\mu\) has mass, the density is the ratio of the point masses: \(f(1)=\tfrac{1/4}{1/2}=\tfrac12\) , \(f(2)=1\) , \(f(3)=\tfrac12\) . The value \(f(4)\) is arbitrary, since \(\{4\}\) is \(\mu\) -null; that is the “unique up to null sets” of the theorem. One checks \(\nu_a(E)=\sum_{x\in E}f(x)\mu(\{x\})\) on all \(16\) subsets; for instance \(\nu_a(\{1,3\})=\tfrac12\cdot\tfrac12+\tfrac12\cdot\tfrac14=\tfrac38\) . The total masses are \(\nu_a(X)=\tfrac58\) and \(\nu_s(X)=\tfrac38\) .
Example 9.5: Absolute continuity without a density
On the Borel sets of \([0,1]\) let \(\mu\) be counting measure and \(\nu=\lambda\) . The only \(\mu\) -null set is \(\emptyset\) , so \(\lambda\ll\mu\) trivially. Suppose \(\lambda(E)=\int_Ef\,d\mu\) for some \(f\ge0\) . Taking \(E=\{x\}\) gives \(0=\lambda(\{x\})=f(x)\) for every \(x\) , so \(f=0\) and \(\lambda=0\) , which is false. There is no Radon–Nikodym derivative. The hypothesis that fails is \(\sigma\) -finiteness of \(\mu\) : an uncountable set is not a countable union of finite sets.
The derivative notation is justified by rules that look like those of calculus.
Proposition 9.12 (rules for Radon–Nikodym derivatives). Let \(\mu,\nu,\rho\) be \(\sigma\) -finite positive measures on \((X,\mathcal A)\) with \(\nu\ll\mu\) .
(a) For every measurable \(g\ge0\) , \(\displaystyle\int g\,d\nu=\int g\,\frac{d\nu}{d\mu}\,d\mu\) .
(b) If \(\rho\ll\nu\) , then \(\rho\ll\mu\) and \(\dfrac{d\rho}{d\mu}=\dfrac{d\rho}{d\nu}\cdot\dfrac{d\nu}{d\mu}\) \(\mu\) -a.e.
(c) If also \(\mu\ll\nu\) , then \(\dfrac{d\mu}{d\nu}=\Bigl(\dfrac{d\nu}{d\mu}\Bigr)^{-1}\) a.e.
Proof. (a) is the second half of Proposition 4.6. For (b), apply (a) with \(g=\mathbf 1_E\,\frac{d\rho}{d\nu}\) :
So the product is a density of \(\rho\) with respect to \(\mu\) , and by uniqueness it is the derivative. For (c), take \(\rho=\mu\) in (b): \(1=\frac{d\mu}{d\mu}=\frac{d\mu}{d\nu}\cdot\frac{d\nu}{d\mu}\) a.e.
The chain rule is proved by integrating over an arbitrary set, never by cancelling differentials. The cancellation is a way of remembering the result.
Example 9.6: A change of reference measure
On \((0,1]\) let \(\nu\) and \(\rho\) be the measures with densities \(2x\) and \(3x^2\) with respect to \(\lambda\) , so that \(\nu((a,b])=b^2-a^2\) and \(\rho((a,b])=b^3-a^3\) . Both densities are strictly positive, so \(\lambda\ll\rho\) and by Proposition 9.12
Test this against sets. For \(0<a<b\le1\) ,
The two sides agree on the half-open intervals, a \(\pi\) -system generating the Borel sets, so by Theorem 1.17 they agree on all Borel sets, and \(\tfrac2{3x}\) is indeed the derivative. Notice that it is unbounded near \(0\) : relative to \(\rho\) , which is very thin near the origin, the measure \(\nu\) is comparatively heavy there.
For a measure that is not absolutely continuous, take \(\nu=\lambda+\delta_{1/2}\) on \((0,1]\) . With respect to \(\lambda\) its Lebesgue decomposition is \(\nu_a=\lambda\) , with density \(1\) , and \(\nu_s=\delta_{1/2}\) , carried by the \(\lambda\) -null set \(\{\tfrac12\}\) .
9.4 Two places the theorem is used
Densities on a coarser family of sets. Let \(\mu\) be a finite measure on \((X,\mathcal A)\) , let \(f\ge0\) be integrable, and let \(\mathcal G\subseteq\mathcal A\) be a smaller \(\sigma\) -algebra. Restricted to \(\mathcal G\) , the measure \(\nu(E)=\int_Ef\,d\mu\) is still absolutely continuous with respect to the restriction of \(\mu\) . The Radon–Nikodym theorem, applied on \((X,\mathcal G)\) , produces a function \(g\) that is measurable with respect to \(\mathcal G\) and satisfies
The function \(f\)
itself will usually not do, because it need not be \(\mathcal G\)
-measurable. This is a genuinely new use of the theorem: no formula for \(g\)
is available in general, and its existence is exactly what the theorem provides. When \(\mu\)
is a probability measure, \(g\)
is the conditional expectation of \(f\)
given \(\mathcal G\)
, the central object of Measure-Theoretic Probability (math.measure-theoretic-probability.conditional-expectation).
Example 9.7: Averaging over the atoms of a coarser sigma-algebra
Let \(X=\{1,2,3,4,5,6\}\) with \(\mu(\{x\})=\tfrac16\) for each point, let \(f(x)=x\) , and let \(\mathcal G\) be the \(\sigma\) -algebra with atoms \(\{1,2\}\) , \(\{3\}\) , \(\{4,5,6\}\) . A \(\mathcal G\) -measurable \(g\) is constant on atoms (Example 3.1), and the defining identity on an atom \(A\) reads \(g\cdot\mu(A)=\int_Af\,d\mu\) . So \(g\) is the average of \(f\) over each atom:
Check on the set \(E=\{1,2,3\}\in\mathcal G\) : \(\int_Eg\,d\mu=\tfrac32\cdot\tfrac26+3\cdot\tfrac16=1\) and \(\int_Ef\,d\mu=\tfrac{1+2+3}6=1\) . The two functions have the same integral over every one of the \(8\) sets of \(\mathcal G\) , and in particular the same total, \(\tfrac{21}6=\tfrac72\) . They do not have the same integral over \(\{1\}\) , which is not in \(\mathcal G\) .
Linear functionals on the Lebesgue spaces. Let \(\mu\) be finite, \(1\le p<\infty\) , and let \(\Lambda:L^p(\mu)\to\mathbb R\) be linear with \(\lvert\Lambda(h)\rvert\le C\lVert h\rVert_p\) . Put \(\nu(E)=\Lambda(\mathbf 1_E)\) . Linearity makes \(\nu\) finitely additive, and the bound makes it countably additive, because the indicators of the tails of a disjoint union tend to \(0\) in \(L^p\) . If \(\mu(E)=0\) then \(\mathbf 1_E\) is the zero element of \(L^p\) , so \(\nu(E)=0\) : thus \(\nu\) is a finite signed measure with \(\nu\ll\mu\) , and it has a density \(g\) . By linearity \(\Lambda(h)=\int hg\,d\mu\) for simple \(h\) , and by density of the simple functions for all \(h\in L^p\) ; a further estimate shows \(g\in L^q\) for the conjugate exponent \(q\) . So every bounded linear functional on \(L^p\) is integration against a function in \(L^q\) , the converse of Proposition 7.11. The details, and what follows from them, belong to Functional Analysis.
The Radon–Nikodym derivative has so far been produced by an existence proof. On the line, with \(\mu=\lambda\) , one would like to compute it as a limit of ratios \(\nu(I)/\lambda(I)\) over shrinking intervals \(I\) , which would justify the word derivative. Does that limit exist, and does it give the density?
Exercises
-
On \(X=\{1,2,3,4,5\}\) let \(\nu\) have the point values \(2,-3,1,0,-1\) . List every Hahn decomposition, and compute \(\nu^+(X)\) , \(\nu^-(X)\) , \(\lvert\nu\rvert(X)\) , and the largest and smallest values of \(\nu(E)\) over all subsets \(E\) .
Answer
Positive sets avoid \(2\) and \(5\) ; negative sets avoid \(1\) and \(3\) ; the point \(4\) is null. The Hahn decompositions are \(P=\{1,3\}\) , \(N=\{2,4,5\}\) and \(P=\{1,3,4\}\) , \(N=\{2,5\}\) . Then \(\nu^+(X)=3\) , \(\nu^-(X)=4\) , \(\lvert\nu\rvert(X)=7\) , \(\nu(X)=-1\) . The largest value of \(\nu(E)\) is \(3\) and the smallest is \(-4\) .
-
Let \(\nu=\mu_1-\mu_2\) where \(\mu_1,\mu_2\) are positive measures, one of them finite, with \(\mu_1\perp\mu_2\) . Prove that \(\mu_1=\nu^+\) and \(\mu_2=\nu^-\) , so that the Jordan decomposition is the only decomposition into mutually singular parts.
Solution
Suppose \(\mu_1\) is the finite one; the other case is symmetric. By mutual singularity there is \(A\in\mathcal A\) with \(\mu_2(A)=0\) and \(\mu_1(X\setminus A)=0\) .
The pair \(A\) , \(X\setminus A\) is a Hahn decomposition for \(\nu\) . If \(E\subseteq A\) is measurable then \(\mu_2(E)=0\) , so \(\nu(E)=\mu_1(E)\ge0\) ; thus \(A\) is positive. If \(E\subseteq X\setminus A\) then \(\mu_1(E)=0\) , so \(\nu(E)=-\mu_2(E)\le0\) ; thus \(X\setminus A\) is negative.
Identification. By Theorem 9.5 the variations may be computed from any Hahn decomposition. For every measurable \(E\) ,
\[ \nu^+(E)=\nu(E\cap A)=\mu_1(E\cap A)-\mu_2(E\cap A)=\mu_1(E\cap A)=\mu_1(E), \]where the third equality uses \(\mu_2(A)=0\) and the last uses \(\mu_1(E\setminus A)\le\mu_1(X\setminus A)=0\) . In the same way \(\nu^-(E)=-\nu(E\setminus A)=\mu_2(E\setminus A)=\mu_2(E)\) .
So a decomposition of \(\nu\) into mutually singular positive parts is forced to be the Jordan decomposition. Without singularity there are many decompositions: \(\nu=(\nu^++\rho)-(\nu^-+\rho)\) for any finite positive measure \(\rho\) . The Jordan decomposition is the one with no cancellation, the minimal one.
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Prove that \(\lvert\nu\rvert(E)=\sup\sum_{i=1}^n\lvert\nu(E_i)\rvert\) , the supremum being over all finite partitions of \(E\) into measurable sets \(E_1,\dots,E_n\) .
Solution
Let \(X=P\cup N\) be a Hahn decomposition and let \(E_1,\dots,E_n\) be a measurable partition of \(E\) . For each \(i\) ,
\[ \lvert\nu(E_i)\rvert=\lvert\nu^+(E_i)-\nu^-(E_i)\rvert\le\nu^+(E_i)+\nu^-(E_i)=\lvert\nu\rvert(E_i). \]Summing over \(i\) and using additivity of the positive measure \(\lvert\nu\rvert\) gives \(\sum_i\lvert\nu(E_i)\rvert\le\lvert\nu\rvert(E)\) . So the supremum is at most \(\lvert\nu\rvert(E)\) .
For the reverse inequality use the partition of \(E\) into the two sets \(E\cap P\) and \(E\cap N\) :
\[ \lvert\nu(E\cap P)\rvert+\lvert\nu(E\cap N)\rvert=\nu^+(E)+\nu^-(E)=\lvert\nu\rvert(E). \]So the supremum is attained, by a partition into two pieces. This formula describes the total variation without reference to the Hahn decomposition, and it is the one that generalizes: for the signed measure \(\nu(E)=\int_Ef\,d\mu\) it says that \(\int_E\lvert f\rvert\,d\mu\) is the largest value of \(\sum_i\lvert\int_{E_i}f\,d\mu\rvert\) , attained by cutting \(E\) along the sign of \(f\) .
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Let \(\nu(E)=\int_E(x^2-x)\,dx\) for Borel sets \(E\subseteq[0,2]\) . Find a Hahn decomposition and compute \(\nu^+([0,2])\) , \(\nu^-([0,2])\) , \(\nu([0,2])\) , and \(\lvert\nu\rvert([0,2])\) .
Solution
The density \(f(x)=x^2-x=x(x-1)\) is negative on \((0,1)\) , zero at \(0\) and \(1\) , and positive on \((1,2]\) . By Example 9.2, \(P=\{f\ge0\}=\{0\}\cup[1,2]\) and \(N=(0,1)\) form a Hahn decomposition; the points \(0\) and \(1\) are null and may be moved to either side. The variations have densities \(f^+\) and \(f^-\) :
\[ \nu^+([0,2])=\int_1^2(x^2-x)\,dx=\Bigl[\frac{x^3}3-\frac{x^2}2\Bigr]_1^2=\frac23+\frac16=\frac56,\qquad\nu^-([0,2])=\int_0^1(x-x^2)\,dx=\frac12-\frac13=\frac16 . \]Hence
\[ \nu([0,2])=\frac56-\frac16=\frac23,\qquad\lvert\nu\rvert([0,2])=\frac56+\frac16=1 . \]As a check, \(\int_0^2(x^2-x)\,dx=\tfrac83-2=\tfrac23\) directly. The total variation \(1\) is the integral of \(\lvert x^2-x\rvert\) , and it exceeds \(\lvert\nu([0,2])\rvert\) by twice the negative variation, the amount lost to cancellation.
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For a signed measure \(\nu\) and a positive measure \(\mu\) , prove that the three conditions \(\nu\ll\mu\) ; \(\lvert\nu\rvert\ll\mu\) ; and both \(\nu^+\ll\mu\) and \(\nu^-\ll\mu\) are equivalent.
Hint
If \(\nu\ll\mu\) and \(\mu(E)=0\) , then \(\mu(E\cap P)=\mu(E\cap N)=0\) , so \(\nu^+(E)=\nu(E\cap P)=0\) and \(\nu^-(E)=0\) . The other implications follow from \(\lvert\nu\rvert=\nu^++\nu^-\) and \(\lvert\nu(E)\rvert\le\lvert\nu\rvert(E)\) .
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Prove that if \(\nu\ll\mu\) and \(\nu\perp\mu\) for positive measures \(\mu,\nu\) , then \(\nu=0\) , identifying the set on which each hypothesis is used.
Answer
By singularity there is \(A\) with \(\mu(A)=0\) and \(\nu(X\setminus A)=0\) . By absolute continuity, used on the \(\mu\) -null set \(A\) , \(\nu(A)=0\) . So \(\nu(X)=\nu(A)+\nu(X\setminus A)=0\) .
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On \((0,1)\) let \(\nu(E)=\int_Ex^{-2/3}\,dx\) . Prove that among Borel sets of Lebesgue measure \(\delta\) the interval \((0,\delta)\) has the largest \(\nu\) -measure, and find the largest \(\delta\) for which \(\lambda(E)<\delta\) implies \(\nu(E)<0.3\) .
Solution
Write \(\phi(x)=x^{-2/3}\) , a decreasing function, and \(I=(0,\delta)\) . Let \(E\subseteq(0,1)\) be a Borel set with \(\lambda(E)=\delta\) . Since \(\lambda(E)=\lambda(I)\) , removing the common part gives \(\lambda(E\setminus I)=\lambda(I\setminus E)\) ; call this number \(m\) . On \(E\setminus I\) every point satisfies \(x\ge\delta\) , so \(\phi\le\delta^{-2/3}\) there; on \(I\setminus E\) every point satisfies \(x<\delta\) , so \(\phi>\delta^{-2/3}\) . Therefore
\[ \nu(E\setminus I)\le\delta^{-2/3}m\le\nu(I\setminus E), \]and adding \(\nu(E\cap I)\) to both ends gives \(\nu(E)\le\nu(I)\) . The same comparison shows that a set of measure less than \(\delta\) has \(\nu\) -measure less than \(\nu((0,\delta))\) .
Now \(\nu((0,\delta))=\int_0^\delta x^{-2/3}\,dx=3\delta^{1/3}\) . So the supremum of \(\nu(E)\) over sets with \(\lambda(E)<\delta\) is \(3\delta^{1/3}\) , approached but not attained. The implication “\(\lambda(E)<\delta\) implies \(\nu(E)<0.3\) ” therefore holds exactly when \(3\delta^{1/3}\le0.3\) , that is, when \(\delta\le10^{-3}\) . The largest admissible value is \(\delta=0.001\) .
The exponent matters: the density \(x^{-2/3}\) is more singular than the \(x^{-1/2}\) of Example 9.3, and the \(\delta\) required for a given \(\varepsilon\) shrinks like \(\varepsilon^3\) instead of \(\varepsilon^2\) .
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A student writes: “Lebesgue measure on \([0,1]\) vanishes on every set on which counting measure vanishes, so by the Radon–Nikodym theorem it has a density with respect to counting measure.” State the hypothesis that fails and show directly that no density exists.
Answer
Counting measure on \([0,1]\) is not \(\sigma\) -finite, so Theorem 9.10 does not apply. If \(\lambda(E)=\int_Ef\,d\#=\sum_{x\in E}f(x)\) , then \(E=\{x\}\) gives \(f(x)=\lambda(\{x\})=0\) for every \(x\) , whence \(\lambda(E)=0\) for all \(E\) , which is false.
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On \(\mathbb N\) let \(\mu\) have weights \(2^{-k}\) and \(\nu\) have weights \(2\cdot3^{-k}\) . Compute \(\dfrac{d\nu}{d\mu}\) and \(\dfrac{d\mu}{d\nu}\) , and verify \(\nu(\mathbb N)=\int\dfrac{d\nu}{d\mu}\,d\mu\) by summing the series.
Answer
\(\dfrac{d\nu}{d\mu}(k)=\dfrac{2\cdot3^{-k}}{2^{-k}}=2\bigl(\tfrac23\bigr)^k\) and \(\dfrac{d\mu}{d\nu}(k)=\tfrac12\bigl(\tfrac32\bigr)^k\) , reciprocals as in Proposition 9.12(c). Check: \(\int\frac{d\nu}{d\mu}\,d\mu=\sum_k2(\tfrac23)^k2^{-k}=\sum_k2\cdot3^{-k}=1=\nu(\mathbb N)\) .
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Let \(\mu_F\) be the Lebesgue–Stieltjes measure of Example 2.4. Find its Lebesgue decomposition with respect to \(\lambda\) , give the density of the absolutely continuous part, and compute the total mass of each part.
Solution
Recall \(F(x)=0\) for \(x<0\) , \(F(x)=x^2\) for \(0\le x<1\) , and \(F(x)=2\) for \(x\ge1\) . The function is differentiable except at \(1\) , with \(F'(x)=2x\) on \((0,1)\) and \(F'=0\) elsewhere, and it has a jump of size \(1\) at \(x=1\) . This suggests
\[ \nu_a(E)=\int_{E\cap(0,1)}2x\,dx,\qquad\nu_s=\delta_1 . \]Verification. Both \(\mu_F\) and \(\nu_a+\nu_s\) are Borel measures finite on bounded intervals, so by the uniqueness in Theorem 2.14 it suffices to compare them on half-open intervals \((a,b]\) . If \(0\le a<b<1\) , then \((\nu_a+\nu_s)((a,b])=\int_a^b2x\,dx=b^2-a^2=F(b)-F(a)\) . If \(0\le a<1\le b\) , then \((\nu_a+\nu_s)((a,b])=(1-a^2)+1=2-a^2=F(b)-F(a)\) . The remaining cases, with \(a<0\) or \(a\ge1\) , are checked the same way.
It is the Lebesgue decomposition. The measure \(\nu_a\) has a density, so \(\nu_a\ll\lambda\) ; and \(\nu_s=\delta_1\) is carried by the \(\lambda\) -null set \(\{1\}\) , so \(\nu_s\perp\lambda\) . By uniqueness in Theorem 9.10 this is the decomposition, with \(\dfrac{d\nu_a}{d\lambda}=2x\,\mathbf 1_{(0,1)}(x)\) .
Masses. \(\nu_a(\mathbb R)=\int_0^12x\,dx=1\) and \(\nu_s(\mathbb R)=1\) , adding up to \(\mu_F(\mathbb R)=2\) . The continuous increase of \(F\) accounts for half of the total mass and the jump for the other half.
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Let \(\mu\) be \(\sigma\) -finite and let \(f,g:X\to[0,\infty]\) be measurable with \(\int_Ef\,d\mu=\int_Eg\,d\mu\) for all \(E\in\mathcal A\) . Prove that \(f=g\) a.e., and show that the conclusion can fail when \(\mu\) is not \(\sigma\) -finite.
Solution
Write \(X=\bigcup_jX_j\) with \(\mu(X_j)<\infty\) . For positive integers \(j,k,m\) let
\[ E_{j,k,m}=X_j\cap\{g\le m\}\cap\{f>g+\tfrac1k\}. \]On this set \(g\) is bounded by \(m\) and the measure is finite, so \(\int_{E_{j,k,m}}g\,d\mu\le m\mu(X_j)<\infty\) , and by hypothesis \(\int_{E_{j,k,m}}f\,d\mu\) is the same finite number. Subtracting two finite integrals is legitimate, and since \(f-g>\tfrac1k\) on the set,
\[ 0=\int_{E_{j,k,m}}(f-g)\,d\mu\ge\frac1k\,\mu(E_{j,k,m}). \]So each \(E_{j,k,m}\) is null. A point with \(f(x)>g(x)\) has \(g(x)<\infty\) , hence lies in some \(E_{j,k,m}\) . Therefore \(\{f>g\}\) is a countable union of null sets, and \(f\le g\) a.e. Exchanging the roles of \(f\) and \(g\) gives \(f=g\) a.e.
The care about finiteness is the whole proof: the naive argument “integrate \(f-g\) over \(\{f>g\}\) ” may involve \(\infty-\infty\) . Without \(\sigma\) -finiteness the statement is false. Let \(X\) be a single point \(x\) with \(\mu(\{x\})=\infty\) , \(f=1\) , and \(g=2\) . Then \(\int_Ef\,d\mu=\int_Eg\,d\mu\) for both measurable sets, being \(0\) on \(\emptyset\) and \(\infty\) on \(X\) , but \(f\ne g\) on a set of positive measure.
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On \((0,\infty)\) let \(\nu\) and \(\rho\) have densities \(e^{-x}\) and \(xe^{-x}\) with respect to \(\lambda\) . Find \(\dfrac{d\rho}{d\nu}\) , and verify Proposition 9.12(a) for \(g=\mathbf 1_{(0,1]}\) by computing \(\rho((0,1])\) in two ways.
Answer
By Proposition 9.12, \(\dfrac{d\rho}{d\nu}=\dfrac{xe^{-x}}{e^{-x}}=x\) . Directly, \(\rho((0,1])=\int_0^1xe^{-x}\,dx=1-\tfrac2e\approx0.2642\) . Through \(\nu\) : \(\int\mathbf 1_{(0,1]}\cdot x\,d\nu=\int_0^1x\,e^{-x}\,dx\) , the same integral.
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On \([0,1)\) with Lebesgue measure let \(\mathcal G\) be the \(\sigma\) -algebra with atoms \([0,\tfrac12)\) and \([\tfrac12,1)\) , and let \(f(x)=x^2\) . Find the \(\mathcal G\) -measurable \(g\) with \(\int_Eg\,d\lambda=\int_Ef\,d\lambda\) for \(E\in\mathcal G\) , and compare with Exercise 8.8.
Solution
A \(\mathcal G\) -measurable function is constant on each atom: \(g=c_1\) on \([0,\tfrac12)\) and \(g=c_2\) on \([\tfrac12,1)\) . The \(\sigma\) -algebra \(\mathcal G\) has four members, \(\emptyset\) , the two atoms, and \([0,1)\) , and the required identity for the atoms implies it for the other two by additivity. On the first atom it reads
\[ c_1\cdot\tfrac12=\int_0^{1/2}x^2\,dx=\tfrac1{24},\qquad\text{so } c_1=\tfrac1{12}, \]and on the second
\[ c_2\cdot\tfrac12=\int_{1/2}^1x^2\,dx=\tfrac13-\tfrac1{24}=\tfrac7{24},\qquad\text{so } c_2=\tfrac7{12}. \]Thus \(g=\tfrac1{12}\mathbf 1_{[0,1/2)}+\tfrac7{12}\mathbf 1_{[1/2,1)}\) , the Radon–Nikodym derivative on \(\mathcal G\) of the measure with density \(x^2\) . As a check, \(\int g\,d\lambda=\tfrac1{24}+\tfrac7{24}=\tfrac13=\int_0^1x^2\,dx\) .
This is exactly the step function found in Exercise 8.8 as the best approximation to \(x^2\) in \(L^2\) among functions constant on the two halves. The agreement is general: for square-integrable \(f\) , the density of \(f\,d\mu\) on a coarser \(\sigma\) -algebra is the \(L^2\) -closest function measurable with respect to it. One description comes from Radon–Nikodym and works for all integrable \(f\) ; the other comes from the geometry of \(L^2\) .
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Let \(\mu,\nu\) be finite positive measures with \(\nu\ll\mu\) and \(f=\dfrac{d\nu}{d\mu}\) . Prove that \(\mu\ll\nu\) if and only if \(f>0\) \(\mu\) -almost everywhere.
Hint
Let \(Z=\{f=0\}\) . Then \(\nu(Z)=\int_Zf\,d\mu=0\) , so if \(\mu\ll\nu\) then \(\mu(Z)=0\) . Conversely, if \(f>0\) \(\mu\) -a.e. and \(\nu(E)=\int_Ef\,d\mu=0\) , then \(f\mathbf 1_E=0\) a.e. by Proposition 4.7(b), which forces \(\mu(E)=0\) .
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On the Borel sets of \(\mathbb R\) let \(\mu=\lambda+\delta_0\) and \(\nu=\lambda\) . Show that \(\nu\ll\mu\) and find \(\dfrac{d\nu}{d\mu}\) ; show that \(\mu\) is not absolutely continuous with respect to \(\nu\) and give its Lebesgue decomposition with respect to \(\nu\) .
Answer
If \(\mu(E)=0\) then \(\lambda(E)=0\) , so \(\nu\ll\mu\) , with \(\dfrac{d\nu}{d\mu}=\mathbf 1_{\mathbb R\setminus\{0\}}\) : indeed \(\int_E\mathbf 1_{\{x\ne0\}}\,d\mu=\lambda(E\setminus\{0\})=\lambda(E)\) . But \(\nu(\{0\})=0\) while \(\mu(\{0\})=1\) , so \(\mu\) is not absolutely continuous with respect to \(\nu\) . Its Lebesgue decomposition with respect to \(\nu=\lambda\) is \(\mu_a=\lambda\) , with density \(1\) , and \(\mu_s=\delta_0\) .
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Let \(\nu_1,\nu_2,\dots\) and \(\mu\) be positive measures and \(\nu=\sum_n\nu_n\) . Prove that if every \(\nu_n\perp\mu\) then \(\nu\perp\mu\) , and if every \(\nu_n\ll\mu\) then \(\nu\ll\mu\) .
Hint
For singularity take \(A_n\) with \(\mu(A_n)=0\) and \(\nu_n(X\setminus A_n)=0\) , and let \(A=\bigcup_nA_n\) : then \(\mu(A)=0\) and every \(\nu_n\) vanishes on \(X\setminus A\) . For absolute continuity, \(\mu(E)=0\) gives \(\nu_n(E)=0\) for all \(n\) .
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Let \(\nu\) be a finite signed measure with Hahn decomposition \(X=P\cup N\) . Prove that \(\max\{\nu(E):E\in\mathcal A\}=\nu^+(X)\) and that the maximum is attained exactly at the sets \(E\) with \(\lvert\nu\rvert(E\triangle P)=0\) .
Solution
For any measurable \(E\) ,
\[ \nu(E)=\nu^+(E)-\nu^-(E)\le\nu^+(E)\le\nu^+(X)=\nu(P). \]So \(\nu(P)=\nu^+(X)\) is the largest value, and it is attained at \(E=P\) ; all quantities are finite because \(\nu\) is.
Equality \(\nu(E)=\nu^+(X)\) holds exactly when both inequalities are equalities, that is, when \(\nu^-(E)=0\) and \(\nu^+(E)=\nu^+(X)\) , the latter being equivalent to \(\nu^+(X\setminus E)=0\) by finiteness. Now compute the total variation of the symmetric difference. Since \(E\triangle P=(E\setminus P)\cup(P\setminus E)\) , with \(E\setminus P\subseteq N\) and \(P\setminus E\subseteq P\) ,
\[ \lvert\nu\rvert(E\triangle P)=\nu^-(E\setminus P)+\nu^+(P\setminus E)=\nu^-(E)+\nu^+(X\setminus E), \]using that \(\nu^-\) vanishes on subsets of \(P\) and \(\nu^+\) on subsets of \(N\) . Both terms are nonnegative, so the sum is \(0\) exactly when \(\nu^-(E)=0\) and \(\nu^+(X\setminus E)=0\) . This is the general form of what Example 9.1 showed on five points: the maximizers of \(\nu\) are the positive sets of Hahn decompositions, and they differ from one another by null sets.