## 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 ${\nu_{a} - \nu_{a}' = 0 = \nu_{s}' - \nu_{s}}$. $\Box$