Let \((X,\mathcal M)\) be a mesurable space. A signed measure on \((X,\mathcal M)\) is a function \(\nu:M \to [-\infty,\infty]\) such that
\(\nu(\varnothing)=0\),
\(\nu\) assumes at most one of the values \(\pm \infty\),
If \((E_n)_{n \in \mathbb{N}}\) is a sequence of disjoint sets in \(\mathcal M\), then \[\nu\Big(\bigcup_{n=1}^\infty E_n\Big)=\sum_{n=1}^{\infty}\nu(E_n),\] where the latter sum converges absolutely if \(\nu\left(\bigcup_{n=1}^{\infty}E_n\right)\) is finite.
Thus every measure is a signed measure; for emphasis we shall sometimes refer to measures as positive measures.
Example 1. If \(\mu_1\), \(\mu_2\) are measures on \((X, \mathcal M)\) and at least one of them is finite then \(\nu=\mu_1 - \mu_2\) is a signed measure.
Example 2. If \(\mu\) is a measure on \((X, \mathcal M)\) and \(f:X \to [-\infty,\infty]\) is a measurable function such that at least one of the integrals \[\int_X f^+d\mu \quad \text{ and }\quad \int_X f^{-}d\mu\] is finite (in this case we call \(f\) an extended \(\mu\)-integrable function), then the function defined by \[\nu(E)=\int_E fd\mu\] is a signed measure.
Remark. As we shall see later every signed measure can be represented in either of these two forms from the previous two examples.
Proof. The same as for positive measures.$$\tag*{$\blacksquare$}$$
Let \(\nu\) be a signed measure on \((X,\mathcal M)\).
A set \(E \in \mathcal M\) is called positive for \(\nu\) if \(\nu(F) \geq 0\) for all \(F \in \mathcal M\) such that \(F \subseteq E\).
A set \(E \in \mathcal M\) is called negative for \(\nu\) if \(\nu(F) \le 0\) for all \(F \in \mathcal M\) such that \(F \subseteq E\).
A set \(E \in \mathcal M\) is called null for \(\nu\) if \(\nu(F)=0\) for all \(F \in \mathcal M\) and \(F \subseteq E\).
Remark. If \(\nu(E)=\int_E fd\mu\), then
\(E\) is positive for \(\nu\) if \(f \geq 0\) \(\mu\)-a.e. on \(E\),
\(E\) is negative for \(\nu\) if \(f \leq 0\) \(\mu\)-a.e. on \(E\),
\(E\) is null for \(\nu\) if \(f = 0\) \(\mu\)-a.e. on \(E\).
Any measurable subset of a positive set is positive and the union of any countable family of positive sets is positive.
Proof. The first part is clear from the definition of positivity. If \(P_1, P_2, \ldots\) are positive sets, let \[Q_n=P_n \setminus \bigcup_{j=1}^{n-1}P_j,\] then \(Q_n \subseteq P_n\) so \(Q_n\) is positive. Hence if \[E \subseteq \bigcup_{j=1}^\infty P_j\] then \[\qquad \qquad \nu(E)=\sum_{n=1}^{\infty}\nu(Q_n \cap E) \geq 0. \qquad \qquad \tag*{$\blacksquare$}\]
Problem. Given a signed measure \(\nu\) on \((X, \mathcal M)\) it is always possible to find a (positive) measure \(\mu\) that dominates \(\nu\), in the sense that \[\nu(E)\le \mu(E) \quad \text{ for all } \quad E\in \mathcal M\] and in addition \(\mu\) is the “smallest” one that has this property.
Let \(\nu\) be a signed measure on \((X,\mathcal M)\). For every \(E\in \mathcal M\) we define a function \(|\nu|\), called the total variation of \(\nu\), by setting \[|\nu|(E)=\sup\Big\{\sum_{k=1}^{\infty}|\nu(E_k)|: (E_k)_{k \in \mathbb{N}}\subseteq \mathcal M \text{ is disjoint and } E=\bigcup_{k \in \mathbb{N}}E_k\Big\}.\] The supremum is taken over all measurable partitions \((E_k)_{k \in \mathbb{N}}\) of \(E\in \mathcal M\).
The total variation \(|\mu|\) of a signed measure \(\mu\) on \((X, \mathcal M)\) is itself a positive measure on \((X,\mathcal M)\) and \(\mu\le |\mu|\).
Proof. Let \((E_j)_{j \in \mathbb{N}} \subseteq \mathcal M\) be a partition of \(E \in \mathcal M\). Let \(t_j<|\mu|(E_j)\). Then each \(E_j\) has a partition \((A_{ij})_{i\in\mathbb N}\) so that \[\sum_{j \in \mathbb{N}}|\mu(A_{ij})|>t_i \quad \text{ for all } \quad i \in \mathbb{N}.\] Since \((A_{ij})_{i,j\in\mathbb N}\) is a partition of \(E\) it follows that \[\sum_{i \in \mathbb{N}}t_i \leq \sum_{i,j \in \mathbb{N}}|\mu(A_{ij})| \leq |\mu|(E).\] Taking the \(\sup\) over all possible choices of \((t_i)_{i \in \mathbb{N}}\) on the left-hand-side of this inequality we obtain that \(\sum_{i \in \mathbb{N}}|\mu|(E_i) \leq |\mu|(E)\).
To prove the opposite inequality let \((A_j)_{j \in \mathbb{N}}\) be any partition of \(E \in \mathcal M\). Then for any fixed \(j \in \mathbb{N}\), the sequence \((A_j \cap E_i)_{i\in \mathbb N}\) is a partition of \(A_j\) and for any \(i \in \mathbb{N}\) the sequence \((A_j \cap E_i)_{j \in \mathbb{N}}\) is a partition of \(E_i\). Hence, \[\begin{aligned} \sum_{j \in \mathbb{N}}|\mu(A_j)|&=\sum_{j \in \mathbb{N}}\Big|\sum_{i \in \mathbb{N}}\mu(A_j \cap E_i)\Big| \leq \sum_{j \in \mathbb{N}}\sum_{i \in \mathbb{N}}|\mu(A_j \cap E_i)|\\&=\sum_{i \in \mathbb{N}}\sum_{j \in \mathbb{N}}|\mu(A_j \cap E_i)| \leq \sum_{i \in \mathbb{N}}|\mu|(E_i), \end{aligned}\] thus \[|\mu|(E) \leq \sum_{j \in \mathbb{N}}|\mu|(E_j),\] since \((A_j)_{j \in \mathbb{N}}\) was an arbitrary partition of \(E\).$$\tag*{$\blacksquare$}$$
Given a signed measure \(\nu\) on \((X, \mathcal M)\) it is now possible to write \(\nu\) as the difference of two (positive) measures. To see this, we define the positive variation and negative variation of \(\nu\) by \[\nu^+=\frac{1}{2}(|\nu|+\nu) \quad \text{ and } \quad \nu^-=\frac{1}{2}(|\nu|-\nu).\]
Remark.
We see that \(\nu^+\) and \(\nu^-\) are positive measures, and \[\nu=\nu^+-\nu^- \quad \text{ and } \quad |\nu|=\nu^++\nu^-.\]
If \(\nu(E)=\infty\) for some \(E\in \mathcal M\), then \(|\nu|(E)=\infty\), and \(\nu^-(E)\) is defined to be \(0\) in this case.
One more piece of terminology: a signed measure \(\nu\) is called finite (resp. \(\sigma\)-finite) if \(|\nu|\) is finite (resp. \(\sigma\)-finite).
Integration with respect to a signed measure \(\nu\) on \((X, \mathcal M)\) is defined by setting \(L^1(\nu)=L^1(\nu^+) \cap L^1(\nu^-)\) and \(\int_X fd\nu =\int_X f d\nu^+-\int_X fd\nu^-.\)
Let \(\mu\) be a positive or signed measure on \((X, \mathcal M)\). We say that \(\mu\) is supported on \(A\in \mathcal M\) if \(\mu(E)=\mu(E\cap A)\) for every \(E\in \mathcal M\).
Equivalently \(\mu(E)=0\) for every \(E\in \mathcal M\) such that \(E\cap A=\varnothing\).
We say that two signed measures \(\mu\) and \(\nu\) on \((X,\mathcal M)\) are mutually singular or that \(\nu\) is singular with respect to \(\mu\) or vice versa if there exist \(E,F \in \mathcal M\) such that \(E \cap F=\varnothing\) and \(\mu\) is supported on \(E\) and \(\nu\) is supported on \(F\). We write that \(\mu \perp \nu.\)
Let \(\nu\) be a signed measure and \(\mu\) be a positive measure on \((X,\mathcal M)\). We say that \(\nu\) is absolutely continuous with respect to \(\mu\) and write \[\nu \ll \mu\] if \(\nu(E)=0\) for every \(E \in \mathcal M\) for which \(\mu(E)=0\).
Remark. One can easily verify that \(\nu \ll \mu\) iff \(|\nu| \ll \mu\) iff \(\nu^+ \ll \mu\) and \(\nu^- \ll \mu\).
Proof. Suppose that \(\nu \ll \mu\) and take \(E\in \mathcal M\) such that \(\mu(E)=0\). Take any disjoint partition \((E_n)_{n\in\mathbb N}\subseteq \mathcal M\) of \(E\) and note that \(\mu(E_n)=0\) for any \(n\in\mathbb N\), thus \(\sum_{n\in\mathbb N}|\nu(E_n)|=0\), since \(\nu \ll \mu\) and consequently \(|\nu|(E)=0\). If \(|\nu| \ll \mu\), then \(\nu \ll \mu\), since \(\nu\le |\nu|\). Finally, \(|\nu| \ll \mu\) iff \(\nu^+ \ll \mu\) and \(\nu^- \ll \mu\), since \(|\nu|=\nu^++\nu^-\). $$\tag*{$\blacksquare$}$$
Remark. Let \(\nu\) be a signed measure and \(\mu\) be a positive measure on \((X,\mathcal M)\). If \(\nu \perp \mu\) and \(\nu \ll \mu\), then \(\nu=0\).
Proof. Let \(E, F\in \mathcal M\) be such that \(E \cap F=\varnothing\), and \(\nu\) is supported on \(E\) and \(\mu\) is supported on \(F\). Then \(\mu(E)=0\), and \(|\nu|(E)=0\), since \(\nu \ll \mu\) iff \(|\nu| \ll \mu\). Hence \(|\nu|(E)=0\), and consequently \(|\nu|=0\) and \(\nu=0\).$$\tag*{$\blacksquare$}$$
Let \(\nu\) be a finite signed measure and \(\mu\) a positive measure on \((X,\mathcal M)\), then \(\nu \ll \mu\) iff for every \(\varepsilon>0\) there exists \(\delta>0\) such that \[\begin{aligned} {\color{red}\mu(E)<\delta \quad \implies \quad |\nu|(E)<\varepsilon.} \end{aligned}\]
Proof (\(\Longleftarrow\)). Since \(\nu \ll \mu\) iff \(|\nu| \ll \mu\) and \(|\nu(E)| \leq |\nu|(E)\), so it suffices to assume that \(\nu=|\nu|\) is positive. If \(E\in \mathcal M\) and \(\mu(E)=0\), then the \(\varepsilon\)-\(\delta\) condition implies that \(\nu \ll \mu\). $$\tag*{$\blacksquare$}$$
Proof (\(\Longrightarrow\)). On the other hand if the \(\varepsilon\)-\(\delta\) condition is not satisfied, there exists \(\varepsilon>0\) such that for all \(n \in \mathbb{N}\) we can find \(E_n \in \mathcal M\) with \[\mu(E_n)<2^{-n} \quad \text{ and } \quad \nu(E_n) \geq \varepsilon.\] Let \(F_k=\bigcup_{j=k}^{\infty}E_j\) and consider \[F=\bigcap_{k=1}^{\infty}F_k.\] Then \(\mu(F)=0\), since \[\mu(F_k)<\sum_{j=k}^{\infty}2^{-j}=2^{1-k}.\] Since \(\nu\) is finite and \(\nu(F_k) \geq \varepsilon\) for all \(k \in \mathbb{N}\), we have \[\nu(F)=\lim_{k \to \infty}\nu(F_k) \geq \varepsilon>0,\] which is impossible, since we have assumed that \(\nu \ll \mu\).$$\tag*{$\blacksquare$}$$
Let \((X, \mathcal M, \mu)\) be a measure space. If \(\mu\) is a measure and \(f\) is an extended \(\mu\)-integrable function, the signed measure \(\nu\) defined by \[\nu(E)=\int_E fd\mu \quad \text{ for } \quad E\in \mathcal M\] is clearly absolutely continuous with respect to \(\mu\). It is also finite iff \(f \in L^1(X, \mu)\). For any complex-valued \(f \in L^1(X, \mu)\) the \(\varepsilon\)-\(\delta\) criterion can be applied to \({\rm Re}(f)\) and \({\rm Im}(f)\) and we obtain the following corollary.
Let \((X, \mathcal M, \mu)\) be a measure space. If \(f \in L^1(X, \mu)\), then for every \(\varepsilon>0\) there is \(\delta>0\) such that \[\mu(E)<\delta \quad \implies \quad \Big|\int_E fd\mu\Big| <\varepsilon.\] We will write \({\color{red}d\nu= fd\mu}\) to express the relationship \({\color{purple}\nu(E)=\int_E fd\mu}.\)
Proposition. Suppose that \((X,\mathcal M, \mu)\) is a finite measure space \(\mu(X)<\infty\). Let \(f \in L^1(X)\) and \(S\subseteq \mathbb C\) be a closed set and \[A_Ef=\frac{1}{\mu(E)}\int_Efd\mu \in S\] for every \(E \in \mathcal M\) with \(\mu(E)>0\). Then \(f(x) \in S\) for almost all \(x \in X\).
Proof. Note that \(S^c=\bigcup_{n\in\mathbb N}B(\alpha_n, r_n)\), where \(B(\alpha_n, r_n)\subseteq S^c\) is a closed disc centered at \(\alpha_n\in\mathbb C\), with radius \(r_n > 0\). Then it suffices to prove that \(\mu(E_n)=0\), with \(E_n=f^{-1}[B(\alpha_n, r_n)]\) for \(n\in\mathbb N\). If we had \(\mu(E_n)>0\) then \[|A_{E_n}(f)-\alpha_n|=\frac{1}{\mu(E_n)}\left|\int_{E_n} (f-\alpha_n)d\mu\right| \leq \frac{1}{\mu(E_n)}\int_{E_n} |f-\alpha_n|d\mu \leq r_n\] this is impossible since \(A_E(f) \in S\). Hence \(\mu(E)=0\).$$\tag*{$\blacksquare$}$$
Let \(\nu\) be a \(\sigma\)-finite signed measure and \(\mu\) be a \(\sigma\)-finite positive measure on \((X,\mathcal M)\). Then the following are true.
(Lebesgue decomposition) There exist unique \(\sigma\)-finite signed measures \(\nu_a\) and \(\nu_s\) on \((X,\mathcal M)\) such that \[\nu_a\ll \mu \quad \text{ and } \quad \nu_s\perp \mu \quad \text{ and } \quad \nu=\nu_a+\nu_s.\]
(Radon–Nikodym theorem) Moreover, there is an extended \(\mu\)-measurable function \(f:X \to \mathbb{R}\) such that \[\nu_a(E)=\int_Ef(x) d\mu(x) \quad \text{ for } \quad E\in\mathcal M\] and any two such functions are equal \(\mu\)-a.e. on \(X\).
Proof of the uniqueness. Suppose that we have another pair of \(\sigma\)-finite signed measures \(\lambda_a\) and \(\lambda_s\) on \((X,\mathcal M)\) such that \[\lambda_a\ll \mu \quad \text{ and } \quad \lambda_s\perp \mu \quad \text{ and } \quad \nu=\lambda_a+\lambda_s.\] Then \[\nu_a+\nu_s=\lambda_a+\lambda_s \quad \iff\quad \lambda=\nu_a-\lambda_a=\lambda_s-\nu_s.\] But \(\lambda=\nu_a-\lambda_a\ll \mu\) and \(\lambda=\lambda_s-\nu_s\perp \mu\), hence we must have \(\lambda=0\), which means that \(\nu_a=\lambda_a\) and \(\lambda_s=\nu_s\) and proves the uniqueness. $$\tag*{$\blacksquare$}$$
Proof of the existence. We shall present von Neumann’s argument that exploits elegantly simple Hilbert space ideas.
We start with the case when both \(\nu\) and \(\mu\) are positive and finite measures. Let \(\lambda=\nu+\mu\) and consider \[\Lambda(\varphi)=\int_X\varphi(x)d\nu(x) \quad \text{ for } \quad \varphi\in L^2(X, \lambda).\]
By the Cauchy–Schwarz inequality we see that the mapping \(\Lambda\) defines a bounded linear functional on \(L^2(X, \lambda)\), namely \[\begin{aligned} |\Lambda(\varphi)|\le \int_X|\varphi(x)|d\nu(x)\le \int_X|\varphi(x)|d\lambda(x)\le \lambda(X)^{1/2}\|\varphi\|_{L^2(X, \lambda)}. \end{aligned}\]
By the Riesz representation theorem, since \(L^2(X, \lambda)\) is a Hilbert space, we find \(g\in L^2(X, \lambda)\) such that \[\Lambda(\varphi)=\int_X\varphi(x)d\nu(x)= \int_X\varphi(x)g(x)d\lambda(x) \quad \text{ for all } \quad \varphi\in L^2(X, \lambda).\]
If \(E\in \mathcal M\) and \(\lambda(E)>0\) and we take \(\varphi=\mathbf{1}_{{E}}\) in the previous identity, and using \(\nu\le \lambda\) we obtain \[0\le \frac{\Lambda(\mathbf{1}_{{E}})}{\lambda(E)}=\frac{1}{\lambda(E)}\int_Eg(x)d\lambda(x)\le 1,\] which by the previous proposition gives \(0\le g(x)\le 1\) for \(\lambda\)-a.e \(x\in X\).
In fact, we can assume that \(0\le g(x)\le 1\) for every \(x\in X\) without disturbing the identity \[\int_X\varphi d\nu= \int_X\varphi g d\lambda \quad \text{ for all } \quad \varphi\in L^2(X, \lambda),\] which can be rewritten in the following form \[\int_X\varphi(1-g)d\nu= \int_X\varphi gd\mu \quad \text{ for all } \quad \varphi\in L^2(X, \lambda). \qquad {\color{purple}(*)}\]
Consider now the two sets \[A=\{x\in X: 0\le g(x)<1\}, \quad \text{ and } \quad B=\{x\in X: g(x)=1\},\] and define two measures \(\nu_a\) and \(\nu_s\) on \(\mathcal M\) by \[\nu_a(E)=\nu(A\cap E) \quad \text{ and } \quad \nu_s(E)=\nu(B\cap E) \quad \text{ for } \quad E\in \mathcal M.\]
To see why \(\nu_s\perp \mu\) we set \(\varphi=\mathbf{1}_{{B}}\) in (*) and obtain \[0=\int_B(1-g)d\nu= \int_Bgd\mu=\mu(B).\]
Finally, we set \(\varphi=\mathbf{1}_{{E}}(1+g+\ldots+g^{n})\), in (*), then we obtain \[\begin{aligned} \int_E(1-g^{n+1})d\nu&=\int_E(1+g+\ldots+g^{n})(1-g)d\nu\\ &= \int_E g(1+g+\ldots+g^{n})d\mu \end{aligned}\]
Since \(1-g^{n+1}(x)=0\) if \(x\in B\) and \(1-g^{n+1}(x) \ _{\overrightarrow{n\to \infty}} \ 1\) for \(x\in A\), the (DCT) implies that \[\lim_{n\to \infty}\int_E(1-g^{n+1})d\nu=\nu(A\cap E)=\nu_a(E).\]
On the other hand, by the (MCT) we obtain \[\nu_a(E)=\lim_{n\to \infty}\int_E g(1+g+\ldots+g^{n})d\mu=\int_Ef d\nu, \ \ \text{ where } \ \ f=\frac{g}{1-g}.\]
Observe also that \(f\in L^1(X, \mu)\), since \(\nu_a(X)\le \nu(X)<\infty\).
If \(\mu\) and \(\nu\) are \(\sigma\)-finite and positive we may clearly find a sequence \((E_j)_{j\in \mathbb N}\subseteq \mathcal M\) of disjoint sets such that \(X=\bigcup_{j\in \mathbb N}E_j\) and \[\mu(E_j)<\infty \quad \text{ and } \quad \nu(E_j)<\infty \quad \text{ for all } \quad j\in \mathbb N.\]
For \(E\in \mathcal M\) we may define positive and finite measures by \[\mu_j(E)=\mu(E\cap E_j) \quad \text{ and } \quad \nu_j(E)=\nu(E\cap E_j) \quad \text{ for all } \quad j\in \mathbb N.\]
Then for each \(j\in\mathbb N\), by the previous part, we can write \[\nu_{j}=\nu_{j, a}+\nu_{j, s}, \quad \text{ where } \quad \nu_{j, a} \ll \mu_j, \quad \text{ and } \quad \nu_{j, s} \perp \mu_j.\]
Moreover, we have \(\nu_{j, a}=f_jd\mu_j\) for some \(f_j\in L^1(X, \mu_j)\).
Then it suffices to set \[f=\sum_{j\in \mathbb N}f_j, \quad \text{ and } \quad \nu_a=\sum_{j\in \mathbb N}\nu_{j, a}, \quad \text{ and } \quad \nu_s=\sum_{j\in \mathbb N}\nu_{j, s}.\]
Finally, if \(\nu\) is signed, then we apply the argument separately to the positive and negative variations of \(\nu\).
This completes the proof of the Lebesgue–Radon–Nikodym theorem. $$\tag*{$\blacksquare$}$$
Let \((X,\mathcal M)\) be a mesurable space. A complex measure on \((X,\mathcal M)\) is a function \(\nu:\mathcal M \to \mathbb C\) such that
\(\nu(\varnothing)=0\),
If \((E_n)_{n \in \mathbb{N}}\) is a sequence of disjoint sets in \(\mathcal M\), then \[\nu\Big(\bigcup_{n=1}^\infty E_n\Big)=\sum_{n=1}^{\infty}\nu(E_n),\] where the series converges absolutely.
Remarks.
Infinite values are not allowed, so a positive measure is a complex measure only if it is finite. For example: if \(\mu\) is positive measure and \(f \in L^1(X, \mu)\) then \(d\nu=fd\mu\) is a complex measure.
If \(\nu\) is a complex measure we shall write \(\nu_r\) and \(\nu_i\) for the real and imaginary parts of \(\nu\).
Thus \(\nu_r\), \(\nu_i\) are signed measures that do not assume values \(\pm \infty\), hence they are finite and so the range of \(\nu\) is a bounded subset of \(\mathbb{C}\).
We define \(L^1(X, \nu)=L^1(X, \nu_r) \cap L^1(X, \nu_i)\) and for \(f \in L^1(X, \nu)\) we set \(\int_X fd\nu=\int_X fd\nu_r+i\int f_Xd\nu_i.\)
If \(\nu\) and \(\mu\) are complex measures we say that \(\nu \perp \mu\) if \(\nu_a \perp \mu_b\) for \(a,b \in \{r,i\}\).
If \(\lambda\) is a positive measure we say \(\nu \ll \lambda\) if \(\nu_r \ll \lambda\) and \(\nu_i \ll \lambda\).
Let \(\nu\) be a complex measure on \((X,\mathcal M)\). For every \(E\in \mathcal M\) we define a function \(|\nu|\), called the total variation of \(\nu\), by setting \[|\nu|(E)=\sup\Big\{\sum_{k=1}^{\infty}|\nu(E_k)|: (E_k)_{k \in \mathbb{N}}\subseteq \mathcal M \text{ is disjoint and } E=\bigcup_{k \in \mathbb{N}}E_k\Big\}.\] The supremum is taken over all measurable partitions \((E_k)_{k \in \mathbb{N}}\) of \(E\in \mathcal M\).
The total variation \(|\nu|\) of a complex measure \(\nu\) on \((X,\mathcal M)\) is a positive measure on \((X, \mathcal M)\).
Proof. The proof goes exactly the same way as for signed measures. $$\tag*{$\blacksquare$}$$
Proposition. Let \(\nu, \nu_1, \nu_2\) be complex measures on \((X,\mathcal M)\), and let \(\mu\) be a positive measure on \((X,\mathcal M)\).
If \(\nu\) is supported on \(A\), so is \(|\nu|\).
If \(\nu_1\perp \nu_2\), then \(|\nu_1|\perp |\nu_2|\).
If \(\nu_1\perp \mu\) and \(\nu_2\perp \mu\), then \(\nu_1+\nu_2\perp \mu\).
If \(\nu_1\ll \mu\) and \(\nu_2\ll \mu\), then \(\nu_1+\nu_2\ll \mu\).
If \(\nu\ll \mu\), then \(|\nu|\ll \mu\).
If \(\nu_1\ll \mu\) and \(\nu_2\perp \mu\), then \(\nu_1\perp \nu_2\).
If \(\nu \ll \mu\) and \(\nu\perp \mu\), then \(\nu=0\).
Proof. It is an easy exercise.$$\tag*{$\blacksquare$}$$
Let \(\nu\) be a complex measure and \(\mu\) be a \(\sigma\)-finite positive measure on \((X,\mathcal M)\). Then the following are true.
(Lebesgue decomposition) There exist unique complex measures \(\nu_a\) and \(\nu_s\) on \((X,\mathcal M)\) such that \[\nu_a\ll \mu \quad \text{ and } \quad \nu_s\perp \mu \quad \text{ and } \quad \nu=\nu_a+\nu_s.\]
(Radon–Nikodym theorem) Moreover, there is a unique \(f\in L^1(X, \mu)\) such that \[\nu_a(E)=\int_Ef(x) d\mu(x) \quad \text{ for } \quad E\in\mathcal M.\]
Proof. It suffices to apply the Lebesgue–Radon–Nikodym theorem for signed measures to the real and imaginary parts of the measure \(\nu\).$$\tag*{$\blacksquare$}$$