Lebesgue Integration of Complex Value Function

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space.Let $f: \Omega \rightarrow \mathbb{C}$ be a complex valued function. It can be written as $f(\omega) \doteq g(\omega) + i h(\omega)$, where $g, h$ are both real functions on $\Omega$.

Definition [Measurability]. A function $f$ is measurable $\mathcal{F}$ if both $g$ and $h$ are.

Definition [Integral]. $f$ is integrable if both $g$ and $h$ are, and its integral is defined by

$$\int f \, d \mu \doteq \int g \, d \mu + i \int h \, d \mu.$$

Theorem. $f$ is integrable if and only if $|f|$ is integrable.

Remark: the integral of the former is complex-valued, while the later is real-valued.

Proof. The claim follows from that

$\square$

Theorem [Linearity]. Let $f_1, f_2$ be integrable. Then $f_1 + f_2$ is integrable, and

$$\int f_1 + f_2 \, d \mu = \int f_1 \, d \mu + \int f_2 \, d \mu.$$

Proof. The integrability of $f_1 + f_2$ follows from the one of $|f_1 + f_2|$, which is integrable as $|f_1 + f_2| \le |f_1| + |f_2|$.

To prove linearity, let $f_1 = g_1 + i h_1$ and $f_2 = g_2 + i h_2$. Then

$\square$

Theorem. If $f$ is integrable, then

$$\left| \int f \, d \mu \right| \le \int |f| \, d \mu.$$

Proof. Let $f = g + i h$. Consider first that $g$ and $h$ are simple functions. W.L.O.G., assume $g = \sum_{k \in [n]} x_k \mathbb{I}_{A_k}$, $h = \sum_{k \in [n]} y_k \mathbb{I}_{A_k}$, where $A_k \in \mathcal{F}, k \in [n]$ constitute a partition of $\Omega$.

For general integrable $f = g + i h$, there exists simple functions $g_n, h_n$ such that

1. $\lim_n g_n = g$ and $\lim_n h_n = h$;

2. $|g_n| \le |g|$ and $|h_n| \le |h|$ for all $n \in \mathbb{N}^+$;

3. $\lim_n \int g_n \, d \mu = \int g \, d \mu$ and $\lim_n \int h_n \, d \mu = \int h \, d \mu$.

It follows that $|g_n + i h_n| \uparrow |f|$. Hence,

where Inequality $(a)$ follows from previous claim, and Inequality $(b)$ follows from monotone convergence theorem.

$\square$

Theorem. Suppose that $f_n = g_n + i h_n$ are complex integrable functions satisfying $\sum_{n \ge 1} \int |f_n| \, d \mu < \infty$. Then $\sum_{n \ge 1} f_n$ is integrable, and

$$\int \sum_{n \ge 1} f_n \, d \mu = \sum_{n \ge 1} \int f_n \, d \mu.$$

Proof. By monotone convergence theorem, we have

Since

$\sum_{n \ge 1} g_n$ is integrable. Via dominated convergence theorem,

Similar argument shows that $\sum_{n \ge 1} h_n$ is integrable, and

Therefore, $\sum_{n \ge 1} f_n$ is integrable, and

$\square$

Reference

[1] P. Billingsley, Probability and measure Anniversary Edition, 3rd ed. New York: Wiley, 1995.