Extending the Radon-Nikodym Theorem to Sigma-Finite Measures

From Finite Measures to Sigma-Finite Measures

The finite Radon–Nikodym theorem gives a density when both measures have finite total mass. When the measures are only sigma-finite, their total masses may be infinite, so that theorem cannot be applied directly to the whole space. The key is to divide the space into countably many measurable pieces on each of which both measures are finite. The finite theorem then supplies a density on each piece, and the disjoint pieces allow those densities to be assembled without ambiguity.

Throughout, $(X,\mathcal{M})$ is a measurable space and $\mu,\nu$ are positive measures on $\mathcal{M}$. A measure is sigma-finite if $X$ is a countable union of measurable sets of finite measure. The notation $\nu\ll\mu$ means that whenever $A\in\mathcal{M}$ and $\mu(A)=0$, then $\nu(A)=0$. No completeness assumption is made. Densities in this tutorial are nonnegative, extended-real-valued measurable functions; equalities between measures hold for every $A\in\mathcal{M}$, while uniqueness of a density is up to $\mu$-almost everywhere equality.

Finite Radon–Nikodym theorem (used here): If $\alpha$ and $\beta$ are finite positive measures on one measurable space and $\beta\ll\alpha$, then there is a nonnegative measurable $h$ such that $\beta(A)=\int_A h\,d\alpha$ for every measurable $A$. The function $h$ is unique $\alpha$-almost everywhere. This is the finite theorem established in the earlier material.

The Sigma-Finite Theorem

Theorem (Radon–Nikodym for sigma-finite measures): Let $(X,\mathcal{M})$ be a measurable space, and let $\mu$ and $\nu$ be sigma-finite positive measures on $\mathcal{M}$. If $\nu\ll\mu$, then there is a nonnegative $\mathcal{M}$-measurable function $f:X\to[0,\infty]$ such that $$ \nu(A)=\int_A f\,d\mu\qquad\text{for every }A\in\mathcal{M}. $$ The function $f$ is unique $\mu$-almost everywhere. The identity allows the value $+\infty$ on either side.

Proof. Sigma-finiteness gives measurable covers $X=\bigcup_{i\ge1}A_i$ and $X=\bigcup_{j\ge1}B_j$, where $\mu(A_i)<\infty$ and $\nu(B_j)<\infty$. The intersections $A_i\cap B_j$ form a countable measurable cover of $X$, and each has finite measure for both measures: $\mu(A_i\cap B_j)\le\mu(A_i)<\infty$ and $\nu(A_i\cap B_j)\le\nu(B_j)<\infty$. Enumerate these intersections as $D_1,D_2,\ldots$, allowing repetitions if necessary, and disjointify them by setting $E_1=D_1$ and $E_n=D_n\setminus\bigcup_{r<n}D_r$ for $n\ge2$. Then the $E_n$ are measurable and pairwise disjoint, their union is $X$, and $\mu(E_n),\nu(E_n)<\infty$ for each $n$. Empty pieces cause no difficulty.

On each $E_n$, use the restricted sigma-algebra $\mathcal{M}|_{E_n}=\{A\cap E_n:A\in\mathcal{M}\}$, and define $\mu_n(A\cap E_n)=\mu(A\cap E_n)$ and $\nu_n(A\cap E_n)=\nu(A\cap E_n)$. These are finite measures. If $\mu_n(C)=0$ for $C\in\mathcal{M}|_{E_n}$, then $\mu(C)=0$, so $\nu(C)=0$ by $\nu\ll\mu$; hence $\nu_n\ll\mu_n$. The finite theorem therefore gives a nonnegative measurable $f_n$ on $E_n$ such that $\nu(A\cap E_n)=\int_{A\cap E_n}f_n\,d\mu$ for every $A\in\mathcal{M}$. Extend each $f_n$ to be zero outside $E_n$, and set $f=\sum_{n=1}^{\infty}f_n$. Each extended function is measurable, so $f$ is measurable as a countable sum of nonnegative measurable functions. Since the pieces are disjoint, at most one term is nonzero at each point.

For any $A\in\mathcal{M}$, countable additivity of $\nu$ on the disjoint sets $A\cap E_n$ gives $\nu(A)=\sum_n\nu(A\cap E_n)$. By the local density identities and the Monotone Convergence Theorem, applied to the increasing partial sums $\sum_{n=1}^N f_n\chi_A$ of nonnegative measurable functions, we have $$ \nu(A)=\sum_{n=1}^{\infty}\int_{A\cap E_n}f_n\,d\mu =\lim_{N\to\infty}\int_A\sum_{n=1}^N f_n\,d\mu =\int_A f\,d\mu. $$ This proves existence, including when the series or integral is infinite.

For uniqueness, suppose $g:X\to[0,\infty]$ is another measurable density for $\nu$. On each $E_n$, both $f$ and $g$ represent the finite measure $\nu_n$ relative to $\mu_n$. Uniqueness in the finite theorem implies $f=g$ $\mu$-almost everywhere on $E_n$. Thus for each $n$ there is a $\mu$-null set $N_n\subseteq E_n$ outside which they agree. The countable union $\bigcup_nN_n$ is null, and outside it $f=g$ on $X=\bigcup_nE_n$. Therefore the density is unique $\mu$-almost everywhere. $\square$

Worked Constructions

Worked Example: A Density on the Whole Real Line

Let $X=\mathbb{R}$, $\mathcal{M}=\mathcal{L}$, and let $\mu=m$ be Lebesgue measure. Define $\nu(A)=2m(A)$ for $A\in\mathcal{L}$. Both measures are sigma-finite, although $\mu(\mathbb{R})=\nu(\mathbb{R})=\infty$. Use the partition $E_k=[k,k+1)$, $k\in\mathbb{Z}$. Each piece has $\mu(E_k)=1$ and $\nu(E_k)=2$, so both restricted measures are finite and the local density is $f_k=2$. The assembled density is $f(x)=2$ for every $x\in\mathbb{R}$.

For a direct numerical check, take $A=[-2,1)\cup[3,5]$. Its intersections with the partition pieces have lengths $1$ for $k=-2,-1,0,3,4$, and length $0$ for all other $k$, up to endpoints of Lebesgue measure zero. Thus $m(A)=1+1+1+1+1=5$, and $\nu(A)=2+2+2+2+2=10$. The assembled density gives $\int_A2\,dm=2\cdot5=10$ as well. In general, the identity follows by summing the local identities over all $k$, not by applying the finite theorem to the infinite-measure space $\mathbb{R}$.

Worked Example: Counting Measure with Unbounded Weights

Let $X=\mathbb{Z}$, let $\mathcal{M}$ be all subsets of $\mathbb{Z}$, and let $\mu$ be counting measure. Define $\nu(A)=\sum_{k\in A}(|k|+1)$, with value $+\infty$ allowed. Both measures are sigma-finite because $X$ is a countable union of singletons and both measures assign finite mass to each singleton. Also $\nu\ll\mu$, since the only set of counting measure zero is the empty set.

Use $E_k=\{k\}$. On this piece $\mu(E_k)=1$ and $\nu(E_k)=|k|+1$, so the local density has value $|k|+1$ at $k$. The assembled function is $f(k)=|k|+1$. For $A=\{-2,0,3\}$, counting measure gives $$ \int_A f\,d\mu=f(-2)+f(0)+f(3)=(2+1)+(0+1)+(3+1)=3+1+4=8. $$ The defining measure formula independently gives $\nu(A)=3+1+4=8$. Also $\mu(A)=3$, so the finite theorem cannot be applied to the whole space: $\mu(\mathbb{Z})=\infty$, and $\nu(\mathbb{Z})=\infty$. The singleton restrictions are finite, which is exactly what the construction uses.

Worked Example: A Density with Infinite Total Integral

Take $X=(0,1]$, with the relative Lebesgue sigma-algebra and $\mu$ equal to Lebesgue measure restricted to $X$. Define $\nu(A)=\int_A x^{-1}\,d\mu$. The partition $E_n=(1/(n+1),1/n]$, $n\ge1$, covers $X$, is disjoint, and has $$ \mu(E_n)=\frac1n-\frac1{n+1}=\frac{1}{n(n+1)},\qquad \nu(E_n)=\int_{1/(n+1)}^{1/n}\frac{dx}{x} =\ln\!\left(\frac{n+1}{n}\right). $$ Both values are finite for every $n$. Thus $\mu$ and $\nu$ are sigma-finite, and $\nu\ll\mu$ because integrating a nonnegative function over a null set gives zero. The local finite-measure density on every piece is $f_n(x)=1/x$; assembling them gives $f(x)=1/x$ on $X$.

For $A=X$, summing the local masses through $N$ pieces gives $$ \sum_{n=1}^N\nu(E_n) =\sum_{n=1}^N\ln\!\left(\frac{n+1}{n}\right) =\ln(N+1), $$ because the product of the fractions $(n+1)/n$ is $N+1$. This tends to $+\infty$. The assembled density gives the same value: $\int_{1/(N+1)}^1x^{-1}\,dx=\ln(N+1)$, and hence $\int_Xx^{-1}\,d\mu=+\infty$. The global identity is valid even though $\nu(X)=\infty$.

Why the Common Partition Matters

A partition chosen only to make $\mu$ finite may leave $\nu$ infinite on individual pieces, so the finite theorem would still not apply there. The proof avoids that problem by first intersecting a finite-measure cover for $\mu$ with a finite-measure cover for $\nu$, and only then making the intersections disjoint. Each resulting piece is finite for both measures. The theorem does not claim that either whole measure is finite; it uses finiteness locally and countable additivity to recover the global identity.

Common mistake. On $X=(0,1]$ in the previous example, one might try to apply the finite theorem directly to $\mu$ and $\nu$, asserting that both are finite because each interval $E_n$ has finite measure. That assertion is wrong: $\mu(X)=1$, but $\nu(X)=\infty$, as the partial sums $\ln(N+1)$ show. The finite theorem's hypotheses concern total mass on the space to which it is applied. The correction is to apply it separately on each $E_n$, where the two masses are $1/(n(n+1))$ and $\ln((n+1)/n)$, and then assemble the densities. This distinction is the essential step from the finite theorem to its sigma-finite version.

Check Your Understanding

For each question, identify the measures and the partition or local identity being used.

Why is $A_i\cap B_j$ finite for both measures when $\mu(A_i)<\infty$ and $\nu(B_j)<\infty$?
Explain why disjointifying a countable cover preserves its union and does not increase the measure of any piece.
In the counting-measure example, compute $\int_{\{-1,2\}}f\,d\mu$ and compare it with $\nu(\{-1,2\})$.
For the density $1/x$ on $(0,1]$, compute $\nu(E_2)$ and $\mu(E_2)$.
Which step in the existence proof uses the Monotone Convergence Theorem, and what nonnegativity condition makes it applicable?