Signed Measures and Lebesgue-Radon-Nikodym Theorem

In this post, we are going to discuss mutually singular and absolutely continuous measures with the goal of proving the Lebesgue-Radon-Nikod\'{y}m theorem.

\bigskip In what follows, {(X,\mathcal{A})} will denote a measurable space. We begin with a review of signed measures.

Definition 1 A signed measure {\nu} on a {\sigma}-algebra {\mathcal{A}} is a set function such that

  1. {\nu} is extended-valued in the sense that {-\infty < \nu(E) \leq \infty} for all {E \in \mathcal{A}}
  2. If {\left\{E_{j}\right\}_{j=1}^{\infty}} are disjoint subsets of {\mathcal{A}}, then

    \displaystyle \nu\left(\bigcup_{j=1}^{\infty}E_{j}\right) = \sum_{j=1}^{\infty}\nu(E_{j})

Note that in order for condition (2) to hold, the sum {\sum \nu(E_{j})} must be independent of any rearrangement. If {\nu(\bigcup_{j=1}^{\infty} E_{j})} is finite, then it follows from a theorem of Riemann that the sum converges absolutely. I have included a proof of this result for the interested reader.

Lemma 2 (Riemann Rearrangement Lemma) Let {\sum a_{n}} be a series of real numbers which converges, but not absolutely. Suppose {-\infty \leq \alpha \leq \beta \leq \infty}. Then there exists a rearrangement {\sum a'_{n}} with partial sums {s'_{n}} such that

\displaystyle \liminf_{n \rightarrow \infty} s'_{n} = \alpha, \indent \limsup_{n \rightarrow \infty} s'_{n} = \beta

Proof: For {n \in \mathbb{N}}, set

\displaystyle p_{n} = \frac{\left|a_{n}\right| + a_{n}}{2}, \indent q_{n} = \frac{\left|a_{n}\right| - a_{n}}{2}

Then {p_{n} - q_{n} = a_{n}, p_{n} + q_{n} = \left|a_{n}\right|, p_{n} \geq 0, q_{n} \geq 0}. I claim that the series {\sum p_{n}, \sum q_{n}} diverge. If either {\sum p_{n}} or {\sum p_{n}} converges, then since {\sum a_{n}} converges, it follows that {\sum \left|a_{n}\right|} converges, which contradicts our hypothesis that {\sum a_{n}} does not converge absolutely.

Let {P_{1},P_{2},\cdots} denote the nonnegative terms of {\sum a_{n}}, in the order in which they occur, and let {Q_{1},Q_{2},\cdots} denote the absolute values of the negative terms of {\sum a_{n}} also in their original order. Since {\sum_{n=1}^{N} P_{n} = \sum_{n=1}^{M_{N}}p_{n}} and {\sum_{n=1}^{N}Q_{n} = \sum_{n=1}^{M_{N}}q_{n}} for all {N \in \mathbb{N}}, both {\sum P_{n}, \sum Q_{n}} diverge.

We will construct sequences {(m_{n})_{n=1}^{\infty}, (k_{n})_{n=1}^{\infty}}, such that

\displaystyle P_{1} + \cdots + P_{m_{1}} - Q_{1} - \cdots - Q_{k_{1}} + P_{m_{1} + 1} + \cdots + P_{m_{2}} - Q_{k_{1} + 1} - \cdots - Q_{k_{2}} + \cdots

satisfies the conclusion of the lemma. Choose real sequences {(\alpha_{n})_{n=1}^{\infty}, (\beta_{n})_{n=1}^{\infty}} such that {\alpha_{n} \rightarrow \alpha, \beta_{n} \rightarrow \beta}. Let {m_{1},k_{1}} be the minimal positive integers such that

\displaystyle P_{1} + \cdots + P_{m_{1}} > \beta_{1},

and

\displaystyle P_{1} + \cdots + P_{m_{1}} - Q_{1} - \cdots - Q_{k_{1}} < \alpha_{1}

Let {m_{2},k_{2}} be the minimal positive integers such that

\displaystyle P_{1} + \cdots + P_{m_{1}} - Q_{1} - \cdots - Q_{k_{1}} + P_{m_{1} + 1} + \cdots + P_{m_{2}} > \beta_{2},

and

\displaystyle P_{1} + \cdots + P_{m_{1}} - Q_{1} - \cdots - Q_{k_{1}} + P_{m_{1} + 1} + \cdots + P_{m_{2}} - Q_{k_{1} + 1} - \cdots - Q_{k_{2}} < \alpha_{2}

We continue in this fashion. This selection process is possible since {\sum P_{n}, \sum Q_{n}} diverge. Let {x_{n},y_{n}} denote the partial sums whose last terms are {P_{m_{n}}} and {-Q_{k_{n}}}, respectively. Then

\displaystyle \left|x_{n} - b_{n}\right| \leq P_{m_{n}}, \indent \left|y_{n} - \alpha_{n}\right| \leq Q_{k_{n}}

Since {\sum a_{n}} is convergent, {P_{m_{n}}, Q_{k_{n}} \rightarrow 0} as {n \rightarrow \infty}. We conclude that {x_{n} \rightarrow \beta, y_{n} \rightarrow \alpha}. Since the subsequential limits of the rearrangement series {\sum a'_{n}} are bounded from above by {\beta} and bounded from below by {\alpha} (this is evident from the squeeze theorem), it follows immediately that

\displaystyle \limsup_{n \rightarrow \infty} s'_{n} = \beta, \indent \liminf_{n \rightarrow \infty} s'_{n} = \alpha

\Box

Corollary 3 If every rearrangement of real-valued series {\sum_{n=1}^{\infty}a_{n}} converges, then {\sum_{n=1}^{\infty}\left|a_{n}\right| < \infty}.

Given a signed measure {\nu} on a measure space {(X,\mathcal{A})}, one might ask if it is always possible to find a positive measure {\mu} which dominates {\nu} in the following sense:

\displaystyle \nu(E) \leq \mu(E), \indent \forall E \in \mathcal{A}

Moreover, one might ask if there is a smallest such {\mu} in the sense that if {\mu'} is any other positive measure which dominates {\nu}, then {\mu(E) \leq \mu'(E)} for all {E \in \mathcal{A}}. It turns out the answer is yes. First, we need to define the notion of the total variation of a measure, analogous to the variation of a measurable function.

Definition 4 Define a set function {\left|\nu\right|: \mathcal{A} \rightarrow \mathbb{R}}, called the total variation of {\nu}, by

\displaystyle \left|\nu\right|(E) = \sup \sum_{j=1}^{\infty}\left|\nu(E_{j})\right|

where the supremum is taken over all countable partitions of {E}.

Lemma 5 The total variation {\left|\nu\right|} of a signed measure {\nu} is a positive measure which satisfies {\nu \leq \left|nu\right|}.

Proof: The only axiom of measures which {\left|\nu\right|} does not obviously satisfy is {\sigma}-additivity. Let {\left\{E_{j}\right\}_{j=1}^{\infty}} be a countable collection of disjoint sets in {\mathcal{A}}, and set {E = \bigcup_{j=1}^{\infty} E_{j}}. For each {j}, let {\alpha_{j} \in \mathbb{R}} such that {\alpha_{j} < \left|v\right|(E_{j})}. It follows from the definition of supremum and of {\left|v\right|} that, for each {j}, there exists a countable collection of disjoint sets {\left\{F_{i,j}\right\}_{i=1}^{\infty}} such that {E_{j} = \bigcup_{i=1}^{\infty}F_{i,j}} and

\displaystyle \alpha_{j} \leq \sum_{i=1}^{\infty}\left|\nu(F_{i,j})\right|

Since {\left\{F_{i,j}\right\}_{i,j=1}^{\infty}} is a partition of {E}, we have

\displaystyle \sum_{j=1}^{\infty}\alpha_{j} \leq \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\left|\nu(F_{i,j})\right| \leq \left|\nu\right|(E)

Taking the supremum over all {\alpha_{j}} which satisfy {\alpha_{j} < \left|v\right|(E_{j})} yields the inequality

\displaystyle \sum_{j=1}^{\infty}\left|\nu\right|(E_{j}) \leq \left|\nu\right|(E)

For the reverse inequality, let {\left\{F_{k}\right\}_{k=1}^{\infty}} be any other partition of {E}. For {k} fixed, {\left\{F_{k} \cap E_{j}\right\}_{j=1}^{\infty}} is a partition of {F_{k}}. By the {\sigma}-additivity of {\nu},

\displaystyle \sum_{k=1}^{\infty}\left|\nu(F_{k})\right| = \sum_{k=1}^{\infty}\left|\sum_{j=1}^{\infty}\nu(F_{k} \cap E_{j})\right|

If {\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}\left|\nu(F_{k} \cap E_{j})\right| = \infty}, then {\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\left|\nu(F_{k} \cap E_{j})\right| = \infty}. It follows from the definition of total variation that {\sum_{j=1}^{\infty}\left|\nu\right|(E_{j}) = \infty}, and the inequality

\displaystyle \sum_{k=1}^{\infty}\left|\nu(F_{k})\right| \leq \sum_{j=1}^{\infty}\left|\nu\right|(E)

holds trivially. If {\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}\left|\nu(F_{k} \cap E_{j})\right| < \infty}, then we can interchange the order of summation to obtaion

\displaystyle \sum_{k=1}^{\infty}\left|\nu(F_{k})\right| = \sum_{k=1}^{\infty}\sum_{j=1}^{\infty}\left|\nu(F_{k} \cap E_{j})\right| = \sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\left|\nu(F_{k} \cap E_{j})\right| \leq \sum_{j=1}^{\infty}\left|\nu\right|(E_{j}),

since {\left\{F_{k} \cap E_{j}\right\}_{k=1}^{\infty}} is a partition of {E_{j}} for each fixed {j}. Since {\left\{F_{k}\right\}_{k=1}^{\infty}} was an arbitrary partition of {E}, we obtain the reverse inequality

\displaystyle \left|\nu\right|(E) \leq \sum_{j=1}^{\infty}\left|\nu\right|(E_{j}),

which completes the proof. \Box

Analogous to the decomposition of a function {f} into a difference {f = f^{+} - f^{-}} of two nonnegative functions, we can write a signed measure as the difference of two positive measures.

Definition 6 For a signed measure {\nu}, we define the positive variation and negative variation of {\nu} by

\displaystyle \nu^{+} = \frac{1}{2}(\left|\nu\right| +\nu) \ \text{and} \ \nu^{-} = \frac{1}{2}(\left|\nu\right| - \nu)

If {E \in \mathcal{A}} is such that {\nu(E) = \infty}, then {\nu^{-}(E) := 0 }. It follows from the preceding proposition that {\nu^{+}, \nu^{-}} are both (positive) measures, and moreover,

\displaystyle \nu = \nu^{+} - \nu^{-} \ \text{and} \ \left|\nu\right| = \nu^{+} + \nu^{-}

We say that the signed measure {\nu} is {\mathbf{\sigma}}-finite if the measure {\left|\nu\right|} is {\sigma}-finite.

Definition 7 Two signed measures {\nu} and {\mu} on a measure space {(X,\mathcal{A})} are mutually singular if there are disjoint subsets {A, B \in \mathcal{A}} so that

\displaystyle \nu(E) = \nu(A \cap E) \ \text{and} \ \mu(E) = \mu(B \cap E), \indent \forall E \in \mathcal{A}

If {\nu,\mu} are mutually singular, we write {\nu \perp \mu}.

If {\nu} is a signed measure and {\nu} is a positive measure on {\mathcal{A}}, we say that {\nu} is absolutely continuous with respect to {\mu} if

\displaystyle E \in \mathcal{A}, \mu(E) = 0 \Rightarrow \nu(E) = 0

If {\nu} is absolutely continuous with respect to {\mu}, we write {\nu \ll \mu}.

Lemma 8 If {\nu} and {\mu} are mutually singular, and {\nu} is also absolutely continuous with respect to {\mu}, then {\nu} vanishes identically.

Proof: Let {A, B \in \mathcal{A}} be disjoint subsets such that

\displaystyle \nu(E) = \nu(A \cap E) \ \text{and} \ \mu(E) = \mu(B \cap E), \indent \forall E \in \mathcal{A}

For any {E \in \mathcal{A}}, {\mu(A \cap E) = 0}. By absolute continuity, {\nu(A \cap E) = \nu(E) = 0}. Since {E \in \mathcal{A}} was arbitrary, we conclude that {\nu = 0}. \Box

Lemma 9 Let {\mu} be a positive measure and {\nu} be a signed measure.

  1. If for every {\epsilon > 0}, there exists {\delta > 0} such that {E \in \mathcal{A}, \mu(E) < \delta \Rightarrow \left|\nu(E)\right| < \epsilon}, then {\nu \ll \mu}.
  2. If {\left|\nu\right|} is a finite measure, then the converse holds.

Proof: We first prove (1). Let {E \in \mathcal{A}} be such that {\mu(E) = 0}. It follows from our hypothesis that {\left|\nu(E)\right| < \epsilon} for every {\epsilon > 0}, from which we conclude that {\nu(E) = 0}.

We prove (2) by contradiction. Suppose there exists {\epsilon > 0} such that for all {\delta > 0}, there exists {E_{\delta} \in \mathcal{A}} with {\mu(E_{\delta}) < \delta} and {\nu(E_{\delta}) \geq \epsilon}. For each {n \in \mathbb{N}} choose {E_{n} \in \mathcal{A}} with {\mu(E_{n}) \leq \frac{1}{2^{n}}} and {\nu(E_{n}) \geq \epsilon}. Then

\displaystyle \mu\left(\limsup_{n \rightarrow \infty} E_{n}\right) = \mu\left(\bigcap_{n=1}^{\infty}\bigcup_{k \geq n} E_{k}\right) \leq \sum_{k=N}^{\infty}\frac{1}{2^{k}}

for all {N \in \mathbb{N}}. We conclude that {\mu(\limsup_{n \rightarrow \infty} E_{n}) = 0}. Since {\nu \ll \mu} by hypothesis, we have {\nu(\limsup_{n \rightarrow \infty} E_{n}) = 0}. But this is a contradiction since

\displaystyle \nu(\limsup_{n \rightarrow \infty} E_{n}) = \limsup_{n \rightarrow \infty} \nu(E_{n}) \geq \epsilon

\Box

There are several proofs of the (Lebesgue-)Radon-Nikodym theorem, but I am fond of the one given below because it uses the theory of Hilbert spaces, which is quite elegant. The exposition closely follows that of Stein and Shakarchi in Real Analysis: Measure, Integration, and Hilbert Spaces.

Theorem 10 Suppose {\mu} is a {\sigma}-finite positive measure on the measure space {(X,\mathcal{A})} and {\nu} is a {\sigma}-finite signed measure on {\mathcal{A}}. Then there exist unique signed meaures {\nu_{a}} and {\nu_{s}} on {mathcal{A}} such that {\nu_{a} \ll \mu, \nu_{s} \perp \mu}, and {\nu = \nu_{s} + \nu_{a}}. Furthermore, the measure {\nu_{a}} is given by

\displaystyle \nu_{a}(E) = \int_{E}f(x)\mu(dx), \indent \forall E \in \mathcal{A}

for some extended {\mu}-integrable function {f}.

Proof: We first consider the case where both {\mu} and {\nu} are positive and finite. Set {\rho = \mu + nu}. We define a functional {\ell: L^{2}(X,\rho) \rightarrow \mathbb{C}} by

\displaystyle \ell(\psi) = \int_{X}\psi(x)\nu(dx)

Clearly, {\ell} is linear. I claim that {\ell} is bounded. Indeed, since {\nu,\mu} are both positive,

\displaystyle \left|\ell(\psi)\right| \leq \int_{X}\left|\psi(x)\right|\nu(dx) \leq \int_{X}\left|\psi(x)\right|\rho(dx) = \int_{X}\left|\psi(x)\mathbf{1}_{X}(x)\right|\rho(dx) \leq (\rho(X))^{\frac{1}{2}}\left(\int_{X}\left|\psi(x)\right|^{2}\rho(dx)\right)^{\frac{1}{2}},

where the last inequality follow from the Cauchy-Schwarz inequality. Since {L^{2}(X,\rho)} is a Hilbert space, the Riesz representation theorem tells us that there exists a unique (up to a.e. equivalence) {g \in L^{2}(X,\rho)} such that

\displaystyle \int_{X}\psi(x)\nu(dx) = \int_{X}\psi(x)\overline{g(x)}\rho(dx), \indent \forall \psi \in L^{2}(X,\rho)

If {E \in \mathcal{A}} with {\rho(E) > 0}, when we set {\psi = \chi_{E}} and recall that {\nu \leq \rho}, we obtain

\displaystyle 0 \leq \frac{1}{\rho(E)}\int_{E}g(x)\rho(dx) = \frac{\nu(E)}{\rho(E)} \leq \frac{\rho(E)}{\rho(E)} = 1

, I claim that {0 \leq g(x) \leq 1} {\rho}-a.e. Indeed, {0 \leq \int_{E}g(x)\rho(dx)} for all sets {E \in \mathcal{A}} implies that

\displaystyle 0 \leq \int_{g < -\frac{1}{n}}g(x)\rho(dx) \leq -\frac{1}{n}\rho\left\{g < -\frac{1}{n}\right\} \Rightarrow \rho\left\{g < -\frac{1}{n}\right\} = 0, \indent \forall n \in \mathbb{N}

Taking the intersection yields {\rho\left\{g < 0\right\} = 0}. By the same argument, {0 \leq \int_{E}(1-g(x))\rho(dx)} for all {E \in \mathcal{A}} implies that {g(x) \leq 1} {\rho}-a.e. Thus, we may assume that {0 \leq g(x) \leq 1} for all {x}, and we have

\displaystyle \int_{X}\psi(1-g)d\nu = \int_{X}\psi g d\mu

Consider the two sets

\displaystyle A = \left\{x \in X: 0 \leq g(x) < 1\right\} \ \text{and} \ B = \left\{x \in X: g(x) = 1\right\}

and define two measures {\nu_{a}} and {\nu_{s}} on {\mathcal{A}} by

\displaystyle \nu_{a}(E) = \nu(A \cap E) \ \text{and} \ \nu_{s}(E) = \nu(B \cap E), \indent \forall E \in \mathcal{A}

The diligent inclined reader can verify that {\nu_{a},\nu_{s}} are indeed measures. I claim that {\nu_{s} \perp \mu}. It is tautological that {A,B} are disjoint and {\nu_{s}(E) = 0} for all measurable subsets {E \subset A}, so we need only show that {\mu(E) = 0} for all measurable subsets {E \subset B}. Indeed, taking {\psi = \mathbf{1}_{E}} in the identity {\int_{X}\psi g d\mu = \int_{X}\psi(1-g)d\nu} yields

\displaystyle \mu(E) = \int_{X}\mathbf{1}_{E}gd\mu = \int_{X}\mathbf{1}_{E}(1-g)d\mu = \int_{X}\mathbf{1}_{E} \cdot 0 d\mu = 0

I now claim that {\nu_{a} \ll \mu}. Let {E \in \mathcal{A}} be such that {\mu(E) = 0}. Then

\displaystyle 0 = \int_{E}gd\mu = \int_{E}(1-g)d\nu

Since {1-g \geq 0}, we conclude that {(1-g)\mathbf{1}_{E} = 0} a.e., which imples that {\nu_{a}(E) = \nu(E \cap A) = 0}. I now claim that {dv_{a} = fd\mu}. Let {E \in \mathcal{A}}, and set {\psi = (\sum_{k=0}^{n}g^{k})\mathbf{1}_{E}}. Then

\displaystyle \int_{E}(1-g^{n+1})d\nu = \int_{E}\sum_{k=1}^{n+1}gd\mu

If {x \in B}, then {(1-g^{n+1})(x) = 0}, and if {x \in A}, then {(1-g^{n+1})(x) \rightarrow 1, n \rightarrow \infty}. In other words, {\lim_{n \rightarrow \infty} 1 - g^{n+1} = \mathbf{1}_{A}}. Since {\left|1-g^{n+1}\right| \leq 2} and our measure space is finite, the dominated convergence theorem implies that

\displaystyle \lim_{n \rightarrow \infty}\int_{E}(1-g^{n+1})d\nu = \int_{E}\mathbf{1}_{A}d\nu = \nu(A \cap E) = \nu_{A}(E)

Observe that

\displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^{n+1}g^{k}(x) = \begin{cases} \frac{g(x)}{1-g(x)} \\ \infty \end{cases}

Set {f = \frac{g}{1-g}}. From the monotone convergence theorem we conclude that

\displaystyle v_{a}(E) = \lim_{n \rightarrow \infty}\int_{E}\sum_{k=1}^{n+1}g^{k}d\mu = \int_{E}fd\mu

Furthermore, {f \in L^{1}(X,\mu)} since {\int_{X}fd\mu = \nu_{a}(X) \leq \nu(X) < \infty}.

We now consider the case where {\nu, \mu} are {\sigma}-finite positive measures. It follows from the definition of {\sigma}-finite that we can find pairwise disjoint sets {E_{j} \in \mathcal{A}} such that {X = \bigcup_{j=1}^{\infty} E_{j}} and {\mu(E_{j}), \nu(E_{j}) < \infty} for all {j}. We define positive, finite measures on {\mathcal{A}} by

\displaystyle \mu_{j}(E) = \mu(E \cap E_{j}) \ \text{and} \ \nu_{j}(E) = \nu(E \cap E_{j}), \indent E \in \mathcal{A}

For each {j}, we write {\nu_{j} = \nu_{j,a} + \nu_{j,s}}, where {\nu_{j,s} \perp \mu_{j}} and {\nu_{j,a} = f_{j}d\mu_{j}}. Defining

\displaystyle f = \sum_{j} f_{j}, \indent \nu_{s} = \sum_{j}\nu_{j,s}, \indent \nu_{a} = \sum_{j}\nu_{j,a}

completes the argument.

If {\nu} is signed, then we apply the preceding argument separately to the positive and negative variations of {\nu}.

To see the uniqueness of the decomposition, suppose we also have {\nu = \nu_{a}' + \nu_{s}'}, where {\nu_{a}' \ll \mu} and {\nu_{s}' \perp \mu}. Then

\displaystyle \nu_{a} - \nu_{a}' = \nu_{s}' - \nu_{s}

Clearly, the LHS is absolutely continuous with respect to {\mu}. I claim that the RHS is singular with respect to {\mu}. Indeed, let {A \coprod B, A' \coprod B'} be paritions guaranteed in the definition of singular measures. Then {\left(\nu_{s}' - \nu_{s}\right)(E) = 0} for all measurable subset {E \subset A \cap A'}, and for any {E \subset X \setminus (A \cap A')},

\displaystyle \mu(E) = \mu(E \cap B \setminus B') + \mu(E \cap B' \setminus B) + \mu(E \cap B \cap B') = 0 + 0 + 0 = 0

We conclude that {\nu_{a} - \nu_{a}' = 0 = \nu_{s}' - \nu_{s}}. \Box

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One Response to Signed Measures and Lebesgue-Radon-Nikodym Theorem

  1. Pingback: Conditional Expectation as a Markov Operator | Math by Matt

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