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/Part 4: Measure Theory and Integration/Chapter 22: The Lebesgue Integral

22.3 Integration of Complex-Valued Functions

Definition 22.3.1 (Integrable).label Let $(X, \cm, \mu)$ be a measure space and $f: X \to \complex$ be a $(\cm, \cb_{\complex})$-measurable function, then $f$ is integrable if

\[\int \abs{f}d\mu < \infty\]

The set

\[\mathcal{L}^{1}(X, \cm, \mu; \complex) = \mathcal{L}^{1}(X, \cm, \mu) = \mathcal{L}^{1}(X; \complex) = \mathcal{L}^{1}(X) = \mathcal{L}^{1}(\mu; \complex) = \mathcal{L}^{1}(\mu)\]

is the vector space of $\mu$-integrable functions on $X$.

Proof. Let $f, g \in \mathcal{L}^{1}(X)$ and $\lambda \in \complex$, then

\[\int \abs{\lambda f + g}d\mu \le \int \abs{\lambda}\abs{f}+ \abs{g}d\mu = \lambda \int \abs{f}d\mu + \int \abs{g}d\mu\]

by Proposition 22.2.7.$\square$

Definition 22.3.2 (Positive and Negative Parts).label Let $X$ be a set and $f: X \to \real$ be a function, then

\[f^{+} = f \vee 0 \quad f^{-} = -(f \wedge 0)\]

are the positive and negative parts of $f$, and $f = f^{+} - f^{-}$.

Definition 22.3.3 (Integral).label Let $(X, \cm, \mu)$ be a measure space and $f \in \mathcal{L}^{1}(X)$. If $f$ is $\real$-valued, then

\[\int f d\mu = \int f^{+} d\mu - \int f^{-} d\mu\]

is the integral of $f$. If $f$ is $\complex$-valued, then

\[\int f d\mu = \int \text{Re}(f)d\mu + i\int \text{Im}(f)d\mu\]

is the integral of $f$.

Proposition 22.3.4 ([Proposition 2.21, 2.22, Fol99]).label Let $(X, \cm, \mu)$ be a measure space, then the integral is a linear functional on $\mathcal{L}^{1}(X)$ such that for any $f \in \mathcal{L}^{1}(X)$, $\abs{\int f d\mu}\le \int \abs{f}d\mu$.

Proof. By Lemma 15.1.13, the mapping

\[I: \mathcal{L}^{1}(X; \real) \to \real \quad f \mapsto \int f^{+} d\mu - \int f^{-} d\mu\]

is a $\real$-linear functional on $\mathcal{L}^{1}(X; \real)$.

Let $f, g \in L^{1}(X; \complex)$ and $\lambda = \alpha + \beta i \in \complex$ with $\alpha, \beta \in \real$, then

\begin{align*}\int (\alpha f)d\mu&= \int \text{Re}(\lambda f)d\mu + i\int \text{Im}(\lambda f)d\mu \\&= \int \alpha \text{Re}(f) - \beta \text{Im}(f)d\mu + i\int \beta\text{Re}(f) + \alpha \text{Im}(f)d\mu \\&= \alpha \braks{\int \text{Re}(f)d\mu + i\int \text{Im}(f)}d\mu + i\beta \braks{\int \text{Re}(f)d\mu + i\int \text{Im}(f)}d\mu \\&= \lambda \int f d\mu\end{align*}

and

\begin{align*}\int f + g d\mu&= \int \text{Re}(f + g)d\mu + i\int \text{Im}(f + g)d\mu \\&= \int \text{Re}(f)d\mu + i\int\text{Im}(f)d\mu + \int \text{Re}(g)d\mu + i\int \text{Im}(g)d\mu \\&= \int f d\mu + \int g d\mu\end{align*}

so the integral is a linear functional on $\mathcal{L}^{1}(X)$.

Finally, if $f$ is $\real$-valued, then

\[\int f = \int f^{+} d\mu - \int f^{-}d\mu \le \int f^{+} + \int f^{-}d\mu = \int \abs{f}d\mu\]

and if $f$ is $\complex$-valued and $\alpha = \ol{\sgn(\int f d\mu)}$, then

\begin{align*}\abs{\int f d\mu}&= \alpha \int f d\mu = \int \alpha f \\&= \text{Re}\paren{\int \alpha f d\mu}= \int \text{Re}(\alpha f)d\mu \\&\le \int \abs{\text{Re}(\alpha f)}d\mu \le \int \abs{\alpha f}d\mu = \int \abs{f}d\mu\end{align*}

so $\abs{\int f d\mu}\le \int \abs{f}d\mu$.$\square$

Remark 22.3.1.label The construction of the Lebesgue integral using positive/negative and real/imaginary parts, and the deduction of properties from this definition both appear cumbersome. This comes from the lack of usage of its linearity on integrable simple functions. A more functional analytic approach will use this linearity and the density of simple functions to construct the integral. This is the route through which the Bochner integral will be presented.

Theorem 22.3.5 (Dominated Convergence Theorem).label Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n}\subset \mathcal{L}^{1}(X)$, and $f:X \to \complex$ such that

  1. (1)

    $f_{n} \to f$ pointwise.

  2. (2)

    There exists $g \in \mathcal{L}^{1}(X)$ such that $\abs{f_n}\le \abs{g}$ for all $n \in \natp$.

then $\int fd\mu = \limv{n}\int f_{n} d\mu$.

Proof [Theorem 2.24, Fol99]. By (1) and (2), $f$ is measurable with $\int \abs{f}d\mu \le \int \abs{g}d\mu < \infty$, so $f \in \mathcal{L}^{1}(X)$. Now, since $g + f, g - f \ge 0$, by Fatou’s lemma and Proposition 22.3.4,

\begin{align*}\int g d\mu + \int f d\mu&\le \liminf_{n \to \infty}\int g + f_{n} d\mu = \int g d\mu + \liminf_{n \to \infty}f_{n} d\mu \\ \int g d\mu - \int f d\mu&\le \liminf_{n \to \infty}\int g - \int f_{n} d\mu = \int g d\mu - \limsup_{n \to \infty}\int f_{n} d\mu\end{align*}

so

\[\limsup_{n \to \infty}\int f_{n} d\mu \le \int f d\mu \le \liminf_{n \to \infty}\int f_{n} d\mu\]

and $\int f d\mu = \limv{n}\int f_{n} d\mu$.$\square$

Remark 22.3.2 (There is no dominated convergence theorem for nets).label In analysis, one frequently encounters places where only sequential continuity is provided or required. It is my opinion that a good portion of this comes from the lack of an extension of the dominated convergence theorem to nets. This limitation arises from the monotone convergence theorem, where continuity from below is used.

For an example, consider the Lebesgue measure on $[0, 1]$. Let $A$ be the net of all finite subsets of $[0, 1]$, directed by inclusion, then $\lim_{\alpha \in A}\one_{\alpha} = 1$ pointwise. However, $\int \one_{\alpha} = 0$ for all $\alpha \in A$.

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