The Radon Nikodym Theorem and Lebesgue Decomposition — Examples

By [L1], $\nu$ is a measure with density $h$ relative to $\lambda$, so in particular $\nu \ll \lambda$.

The measure $\lambda$ is sigma-finite, $\nu$ is finite because $\nu (\mathbb{R})=1$, and $[-n,n]$ is a common finite exhaustion. For every measurable $E$, the definition of $\nu$ gives $\nu (E)={\int }_{E}hd\lambda$. Therefore [L3] identifies $h$ as a representative of $d\nu \mathrm{/}d\lambda$.

The measure $\frac{1}{2}\lambda \text{\hspace{0.17em}⁣}{\upharpoonright }_{[0,1]}$ is absolutely continuous with respect to $\lambda$ by [A1], while $\frac{1}{2}{\mu }_c$ is singular with respect to $\lambda$ by [L1]. Their sum is $\nu$ by definition.

The sets $X_n=[-n,n]$ form a common finite exhaustion for $\lambda$ and the finite measure $\nu$. Since $\nu =\frac{1}{2}\lambda \text{\hspace{0.17em}⁣}{\upharpoonright }_{[0,1]}+\frac{1}{2}{\mu }_c$ is already a decomposition into an absolutely continuous part and a singular part, [L3] forces it to be the Lebesgue decomposition of $\nu$ relative to $\lambda$.

By [L1], ${\delta }_0$ is discrete. By [A1], $\lambda \text{\hspace{0.17em}⁣}{\upharpoonright }_{[0,1]}$ is absolutely continuous with respect to $\lambda$. Their sum is $\mu$, and neither summand has an atomless singular part.

Therefore the displayed decomposition has exactly the form required by [L3], and uniqueness there forces ${\mu }_{\mathrm{d}}={\delta }_0$, ${\mu }_{\mathrm{a}\mathrm{c}}=\lambda \text{\hspace{0.17em}⁣}{\upharpoonright }_{[0,1]}$, and ${\mu }_{\mathrm{s}\mathrm{c}}=0$.

On the measurable space $[0,1]$, the measures $\lambda$, $\mu$, and $\nu$ are finite, their one-set exhaustion is common, and [L3] gives $\nu \ll \mu \ll \lambda$. For every measurable set $E$, one has $\mu (E)={\int }_{E}2d\lambda$ and $\nu (E)={\int }_{E}2x{\chi }_{[0,1]}(x)d\lambda (x)$. Therefore [L1] lets us take $d\mu \mathrm{/}d\lambda =2$ and $d\nu \mathrm{/}d\lambda =2x{\chi }_{[0,1]}$.

The function $x{\chi }_{[0,1]}$ satisfies ${\int }_{E}xd\mu ={\int }_{E}2xd\lambda =\nu (E)$, so it represents $d\nu \mathrm{/}d\mu$. On the measurable space $[0,1]$, all three measures are finite, the one-set exhaustion $X_1=[0,1]$ is common, and $\nu \ll \mu \ll \lambda$. Applying [L2] now yields $\frac{d\nu }{d\lambda }=\frac{d\nu }{d\mu }\frac{d\mu }{d\lambda }=x\cdot 2=2x\lambda \text{-almost everywhere on }[0,1]\mathrm{.}$

The function $F$ is nondecreasing and right-continuous, so [L1] gives a Borel measure ${\mu }_F$. For every $a<b$, direct integration shows ${\int }_{(a,b]}2x{\chi }_{[0,1]}(x)d\lambda (x)=F(b)-F(a),$ because both sides are $0$ off $[0,1]$, and on $[0,1]$ they equal $b^2-a^2$ or the corresponding truncated interval increment.

By [L3], $E\mapsto {\int }_{E}2x{\chi }_{[0,1]}d\lambda$ is a finite Borel measure and hence is finite on compact sets. The Lebesgue--Stieltjes measure ${\mu }_F$ is also finite, because $F$ has total increment $1$. The two measures agree on every half-open interval by step 1.1, so [L1] makes them equal on all Borel sets. Thus [L4] makes ${\mu }_F\ll \lambda$; $\lambda$ is sigma-finite and $[-n,n]$ is a common finite exhaustion. Therefore [L2] identifies $2x{\chi }_{[0,1]}$ as a representative of $d{\mu }_F\mathrm{/}d\lambda$.

Suppose $\lambda (E)={\int }_{E}fdc$ for every measurable set $E\subseteq [0,1]$. Applying this to a singleton $\{x\}$ and using [L1] gives $0=\lambda (\{x\})={\int }_{\{x\}}fdc=f(x)(x\in [0,1])\mathrm{.}$ Thus $f=0$ pointwise.

Step 1.1 and [L2] then give $\lambda ([0,1])={\int }_{[0,1]}fdc=0,$ contradicting the fact that the interval has Lebesgue mass $1$. Therefore no such density exists.

By [L1], the measure $\nu$ is absolutely continuous with respect to $\lambda$.

Let $\epsilon :=1$ and let $\delta >0$. Put $a:=\min (\delta \mathrm{/}2,1\mathrm{/}2)$ and $E:=(0,a)\subseteq (0,1)$. Then $\lambda (E)=a<\delta$, but $\nu (E)={\int }_0^a\frac{dx}{x}=+\mathrm{\infty }>1.$ The measure $\nu$ is sigma-finite: $(0,1)={\bigcup }_{m\ge 2}(1\mathrm{/}m,1)$ and every displayed interval has finite $\nu$-mass, while $\nu$ vanishes off $(0,1)$. Thus the epsilon-delta conclusion fails for an absolutely continuous sigma-finite measure.

Let $h_0:=0$ and $h_1:={\chi }_C$. For every measurable set $E$, [L1] and [L2] give ${\int }_E{h}_{1}d\lambda =\lambda (E\cap C)=0={\int }_E{h}_{0}d\lambda \mathrm{.}$ Thus both $h_0$ and $h_1$ represent the zero measure relative to $\lambda$.

The functions $h_0$ and $h_1$ are not equal pointwise because $h_1=1$ on $C$, but [L3] says only almost-everywhere equality is required. Hence pointwise uniqueness fails.

Suppose $\nu ={\nu }_a+{\nu }_s$ were such a decomposition. For each $x\in [0,1]$, absolute continuity gives ${\nu }_a(\{x\})=0$, so ${\nu }_s(\{x\})=\nu (\{x\})=1.$

If ${\nu }_s\perp \lambda$, then ${\nu }_s$ is concentrated on some Lebesgue-null set $N$. But every $x\in [0,1]\setminus N$ satisfies $\{x\}\subseteq [0,1]\setminus N$, so concentration would force ${\nu }_s(\{x\})=0$, contradicting step 1.1. Therefore no such decomposition exists.

The density $x^{-1\mathrm{/}2}{\chi }_{(0,1]}$ is integrable on $(0,1]$, since ${\int }_0^1{x}^{-1\mathrm{/}2}d\lambda =2.$ Therefore [L1] makes $\nu$ a finite measure, and [L2] shows that it vanishes on every Lebesgue-null set. Thus $\nu \ll \lambda$; the exhaustion $[-n,n]$ verifies the remaining Radon--Nikodym hypotheses.

The same density represents $d\nu \mathrm{/}d\lambda$, but it is unbounded near $0$. Hence a finite absolutely continuous measure need not have a bounded derivative.

The measure $\nu$ is not absolutely continuous with respect to $\lambda$: the Cantor set $C$ is Lebesgue null by [L1], but $\nu (C)=\lambda (C)+{\mu }_c(C)=1.$

By [L2] and [L3], $\lambda ((0,1])=1$, so a Lebesgue-null set $N$ satisfies $\lambda ((0,1]\setminus N)=1$. The measure $\nu$ is therefore not singular with respect to $\lambda$: if it were concentrated on $N$, then $0<\lambda ((0,1]\setminus N)\le \nu ((0,1]\setminus N),$ contradicting concentration on $N$. Thus $\nu$ is neither absolutely continuous nor singular.

The functions $0$ and ${\chi }_C$ both integrate to $0$ over every measurable set by [L1] and [L2], so they both represent the Radon-Nikodym derivative of the zero measure with respect to Lebesgue measure.

These two representing functions are not equal pointwise, while [L3] allows exactly this kind of null-set discrepancy. Hence pointwise uniqueness is false.

By [L1], the measure $\nu$ satisfies $\nu \ll \lambda$.

With $\epsilon =1$, every $\delta >0$ fails: for $E=(0,\delta \mathrm{/}2)$ one has $\lambda (E)<\delta$ but $\nu (E)={\int }_0^{\delta \mathrm{/}2}\frac{dx}{x}=+\mathrm{\infty }>1.$ Thus the epsilon-delta condition does not follow from absolute continuity in this sigma-finite but nonfinite case.

Lebesgue measure is absolutely continuous with respect to counting measure on $[0,1]$, because every counting-null measurable set is empty. Moreover, a set has finite counting measure only when it is finite. If $c$ were sigma-finite, $[0,1]$ would be a countable union of finite sets and hence countable, contradicting [L3]. Thus the dominating measure is not sigma-finite.

If a density $f$ as in [L2] existed, then applying it to singletons and using [L1] would give $f(x)=0$ for every $x\in [0,1]$. But [L4] gives $\lambda ((0,1])=1$, whereas the assumed representation gives $\lambda ((0,1])={\int }_{(0,1]}fdc=0$. Hence the theorem fails without sigma-finiteness of the dominating measure.

The function $x^{-1\mathrm{/}2}{\chi }_{(0,1]}$ defines a finite measure by [L1], and [L2] makes it absolutely continuous with respect to $\lambda$. Because $\lambda$ is sigma-finite and the measure is finite, the exhaustion $[-n,n]$ verifies the remaining Radon--Nikodym hypotheses. Moreover ${\int }_0^1{x}^{-1\mathrm{/}2}d\lambda =2.$

The same function is a representative of $d\nu \mathrm{/}d\lambda$, and it is unbounded near $0$. Therefore absolute continuity alone does not force boundedness of the derivative.