Integrating Functions Against Complex Measures

From Complex Measures to Complex Integrals

A complex measure can assign non-real values to measurable sets, so its integral must preserve both magnitude and phase. The natural control measure is its total variation: it records the size of the measure without cancellation. We first define the integral on simple functions, then extend it to every function integrable with respect to total variation.

Let $(X,\mathcal M)$ be a measurable space, and let $\nu:\mathcal M\to\mathbb C$ be a countably additive complex measure with finite total variation. Its total variation $|\nu|$ is the finite positive measure defined on $E\in\mathcal M$ by taking the supremum of $\sum_{j=1}^k|\nu(E_j)|$ over finite measurable partitions $E=\bigcup_{j=1}^k E_j$. We use the standard properties of total variation: $|\nu(E)|\le|\nu|(E)$, and $|\nu|(X)<\infty$. Thus $(X,\mathcal M,|\nu|)$ is a finite measure space. Functions below are complex-valued and $\mathcal M$-measurable; integrability means $\int |f|\,d|\nu|<\infty$.

There is also a positive-measure description of $\nu$. Write $\nu=\nu_R+i\nu_I$, where $\nu_R$ and $\nu_I$ are the real and imaginary signed measures. Their Jordan decompositions are $\nu_R=\nu_R^+-\nu_R^-$ and $\nu_I=\nu_I^+-\nu_I^-$, with each component a finite positive measure. The standard variation inequalities give $\nu_R^\pm\le|\nu|$ and $\nu_I^\pm\le|\nu|$. These four positive measures let us define the integral using ordinary Lebesgue integrals against positive measures.

Definition: If $\varphi=\sum_{j=1}^k a_j\chi_{E_j}$ is a complex simple function, where the $E_j\in\mathcal M$ are disjoint and $a_j\in\mathbb C$, define $$ \int\varphi\,d\nu =\sum_{j=1}^k a_j\nu(E_j) =\int\varphi\,d\nu_R^+-\int\varphi\,d\nu_R^- +i\int\varphi\,d\nu_I^+-i\int\varphi\,d\nu_I^-. $$ For an $|\nu|$-integrable complex-valued measurable function $f$, define $\int f\,d\nu$ by extending this simple-function integral through approximation in $L^1(|\nu|)$. The definition and the validity of this extension are justified below.

For a complex simple function, integration against a positive measure is defined by linearity from indicators; its real and imaginary parts give the usual complex Lebesgue integral. The component formula above follows from $\nu(E)=\nu_R^+(E)-\nu_R^-(E)+i\nu_I^+(E)-i\nu_I^-(E)$ on every measurable set $E$. It is independent of the chosen disjoint representation: two representations have a common finite measurable refinement, and splitting a set into disjoint pieces preserves the sum of its $\nu$-values by finite additivity.

Worked Example: A Finite Atomic Complex Measure

Let $X=\{1,2,3\}$ with all subsets measurable, and set $\nu(\{1\})=1+i$, $\nu(\{2\})=-2i$, and $\nu(\{3\})=2$. Let $f(1)=2-i$, $f(2)=i$, and $f(3)=-1$. The singleton partition gives $|\nu|(X)=|1+i|+|-2i|+|2|=\sqrt2+2+2=\sqrt2+4$. The integral is

\int f\,d\nu=(2-i)(1+i)+i(-2i)+(-1)(2) =(3+i)+2-2=3+i.

Here $|f(1)|=\sqrt5$, $|f(2)|=1$, and $|f(3)|=1$. Therefore $\int|f|\,d|\nu|=\sqrt5\sqrt2+1\cdot2+1\cdot2=\sqrt{10}+4$. The bound is $|3+i|=\sqrt{10}\le\sqrt{10}+4$. Check: the modulus of the computed integral is $\sqrt{3^2+1^2}=\sqrt{10}$, as used in the inequality.

Extension from Simple Functions

We record why the definition extends to every $f$ with $\int|f|\,d|\nu|<\infty$. Such functions can be approximated in $L^1(|\nu|)$ by complex simple functions. In this setting, the approximation can be seen directly: truncate $f$ to zero on $\{|f|>n\}$, and on $\{|f|\le n\}$ round its real and imaginary parts to a grid with spacing $1/n$. The result $\varphi_n$ is simple, and on $\{|f|\le n\}$ it satisfies $|f-\varphi_n|\le\sqrt2/n$. On the complement, $|f-\varphi_n|=|f|$. Consequently,

\int|f-\varphi_n|\,d|\nu| \le \frac{\sqrt2}{n}|\nu|(X)+\int_{\{|f|>n\}}|f|\,d|\nu|\longrightarrow0.

The last term tends to zero by the Dominated Convergence Theorem: its integrand is bounded by the integrable function $|f|$ and tends pointwise to zero. The first term tends to zero because $|\nu|(X)$ is finite.

Theorem: For every $|\nu|$-integrable complex-valued measurable function $f$, the simple-function definition has a unique extension $\int f\,d\nu$. For $|\nu|$-integrable $f,g$ and $\alpha,\beta\in\mathbb C$, $$ \int(\alpha f+\beta g)\,d\nu =\alpha\int f\,d\nu+\beta\int g\,d\nu, \qquad \left|\int f\,d\nu\right|\le\int|f|\,d|\nu|. $$ Both statements concern finite complex measures and hold for functions identified up to equality $|\nu|$-almost everywhere.

Proof. For a simple function $\psi=\sum_j c_j\chi_{F_j}$ written on disjoint measurable sets, the total-variation inequality gives $$ \left|\int\psi\,d\nu\right| =\left|\sum_jc_j\nu(F_j)\right| \le\sum_j|c_j|\,|\nu(F_j)| \le\sum_j|c_j|\,|\nu|(F_j) =\int|\psi|\,d|\nu|. $$ If $(\varphi_n)$ is any sequence of simple functions converging to $f$ in $L^1(|\nu|)$, then applying this inequality to $\varphi_n-\varphi_m$ shows $$ \left|\int\varphi_n\,d\nu-\int\varphi_m\,d\nu\right| \le\int|\varphi_n-\varphi_m|\,d|\nu|\longrightarrow0 $$ as $m,n\to\infty$. Thus the integrals form a Cauchy sequence in $\mathbb C$, which is complete, and have a limit. If $(\psi_n)$ is another such approximation, then $$ \left|\int\varphi_n\,d\nu-\int\psi_n\,d\nu\right| \le\int|\varphi_n-f|\,d|\nu|+\int|\psi_n-f|\,d|\nu|\longrightarrow0. $$ The limit is therefore independent of the approximation, proving existence and uniqueness of the extension. This construction also agrees with the four positive-measure integrals in the definition: each component is bounded above by $|\nu|$, so convergence in $L^1(|\nu|)$ implies convergence in $L^1$ for every component.

For linearity, choose simple approximations $\varphi_n\to f$ and $\psi_n\to g$ in $L^1(|\nu|)$. Then $\alpha\varphi_n+\beta\psi_n\to\alpha f+\beta g$, since $$ \int|\alpha(\varphi_n-f)+\beta(\psi_n-g)|\,d|\nu| \le|\alpha|\int|\varphi_n-f|\,d|\nu|+|\beta|\int|\psi_n-g|\,d|\nu|\longrightarrow0. $$ Linearity holds for simple functions by finite sums. Taking limits in that identity proves linearity for $f,g$. Finally, for simple $\varphi_n\to f$, the simple-function bound gives $\left|\int\varphi_n\,d\nu\right|\le\int|\varphi_n|\,d|\nu|$. The left side tends to $\left|\int f\,d\nu\right|$, because modulus is continuous. Also, $$ \left|\int|\varphi_n|\,d|\nu|-\int|f|\,d|\nu|\right| \le\int\big||\varphi_n|-|f|\big|\,d|\nu| \le\int|\varphi_n-f|\,d|\nu|\longrightarrow0. $$ The first inequality is the usual integral bound for real integrable functions against the positive measure $|\nu|$; the second is the pointwise reverse triangle inequality. Passing to the limit proves the asserted variation bound. If two functions agree $|\nu|$-a.e., their $L^1(|\nu|)$ distance is zero, so the construction gives the same integral. $\square$

Worked Example: A Complex Density on an Interval

On $([0,1],\mathcal L)$, let $\nu(E)=\int_E(1+ix)\,dm$, and take $f(x)=x$. The density has modulus $\sqrt{1+x^2}$, so $|\nu|(E)=\int_E\sqrt{1+x^2}\,dm$. This is the standard variation formula for a complex measure with an integrable density. In particular, $f$ is $|\nu|$-integrable, since $0\le x\sqrt{1+x^2}\le\sqrt2$ on $[0,1]$. Using the density,

\int f\,d\nu=\int_0^1x(1+ix)\,dx =\int_0^1x\,dx+i\int_0^1x^2\,dx =\frac12+\frac{i}{3}.

The variation integral is exactly $$ \int|f|\,d|\nu|=\int_0^1x\sqrt{1+x^2}\,dx =\left[\frac{(1+x^2)^{3/2}}{3}\right]_0^1 =\frac{2\sqrt2-1}{3}. $$ The left side of the bound is $\left|\frac12+\frac{i}{3}\right|=\frac{\sqrt{13}}6$. To check the inequality exactly, both sides are positive, and their squares satisfy $$ \frac{13}{36}<\frac{(2\sqrt2-1)^2}{9} \quad\Longleftrightarrow\quad 13<36-16\sqrt2 \quad\Longleftrightarrow\quad 16\sqrt2<23. $$ The last inequality follows by squaring positive quantities: $512<529$. Thus the computed values satisfy the bound.

Check: differentiating $(1+x^2)^{3/2}/3$ gives $x\sqrt{1+x^2}$, and evaluating the primitive at $1$ and $0$ gives $(2\sqrt2-1)/3$, as required.

Countable Atoms and a Common Pitfall

Worked Example: Integrating over Countably Many Atoms

Let $X=\mathbb N=\{1,2,\ldots\}$ with all subsets measurable. Set $\nu(\{n\})=(i/2)^n$ and $f(n)=(-1)^n$. Since $\sum_{n=1}^\infty|\nu(\{n\})|=\sum_{n=1}^\infty2^{-n}=1$, this defines a finite complex measure and $\int|f|\,d|\nu|=1$. The integral is the absolutely convergent series

\int f\,d\nu=\sum_{n=1}^\infty(-1)^n(i/2)^n =\sum_{n=1}^\infty(-i/2)^n =\frac{-i/2}{1+i/2} =-\frac15-\frac{2i}{5}.

For the final simplification, multiply numerator and denominator by $1-i/2$: the numerator becomes $(-i/2)(1-i/2)=-1/4-i/2$, and the denominator is $1+1/4=5/4$. Check: multiplying the answer by $1+i/2$ gives $(-1/5-2i/5)(1+i/2)=-i/2$, the first term of the geometric-series formula $r/(1-r)$ with $r=-i/2$. The bound reads $\sqrt5/5\le1$, which holds because $5\le25$.

A frequent error is to replace integration against $\nu$ by integration against $|\nu|$. These are different measures: total variation controls the integral but does not preserve its complex values. For a one-point example, let $X=\{a\}$, $\nu(\{a\})=i$, and $f(a)=1$. Then $|\nu|(\{a\})=1$, so $\int f\,d\nu=i$, whereas $\int f\,d|\nu|=1$. The correct conclusion is the bound $\left|\int f\,d\nu\right|\le\int|f|\,d|\nu|$, not equality with the variation integral.

The positive components explain how the complex integral is assembled; total variation supplies a single positive measure for approximation and estimates. This combination makes the definition stable under $L^1(|\nu|)$ limits and is the basic tool for controlling complex-measure pairings.

Check Your Understanding

For each question, keep the complex integral and the total-variation integral distinct.

For a simple function $\varphi=\sum_{j=1}^k a_j\chi_{E_j}$ on disjoint measurable sets, write $\int\varphi\,d\nu$ using the four positive measures from the Jordan decompositions.
Why does convergence of simple functions to $f$ in $L^1(|\nu|)$ make their complex integrals a Cauchy sequence?
On a one-point space with $\nu(\{a\})=-i$ and $f(a)=2$, compute $\int f\,d\nu$, $|\nu|(\{a\})$, and $\int|f|\,d|\nu|$.
For the countable atomic example, compute the first two terms of the series for $\int f\,d\nu$, and state why the full series converges absolutely.
Explain why changing an integrable function on a $|\nu|$-null set does not change its integral.

Integrating Functions Against Complex Measures

What You'll Learn