16.3 Integration of Complex-Valued Functions

Definition 16.3.1 (Integrable). 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 16.2.7.$\square$

Definition 16.3.2 (Positive and Negative Parts). Let $X$ be a set and $f: X \to \real$ be a function, then

\[f^{+} = \max(f, 0) \quad f^{-} = -\min(f, 0)\]

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

Definition 16.3.3 (Integral). 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 16.3.4. 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. Let $f, g \in \mathcal{L}^{1}(X)$ and $\lambda \in \complex$. First suppose that $f, g$ are $\real$-valued and $\lambda \in \real$. In which case,

\[\int \lambda f d\mu = \int (\lambda f)^{+} d\mu - \int (\lambda f)^{-} d\mu = \begin{cases}\lambda\int f^{+} d\mu - \lambda\int f^{-} d\mu&\lambda \ge 0 \\ -\lambda\int f^{-} d\mu + \lambda\int f^{+} d\mu&\lambda < 0\end{cases}\]

by Proposition 16.2.7, so $\int \lambda f d\mu = \lambda \int f d\mu$.

Let $h = f + g$, then $h = h^{+} - h^{-} = f^{+} + g^{+} - f^{-} - g^{-}$, so

\begin{align*}h^{+} + f^{-} + g^{-}&= h^{-} + f^{+} + g^{+} \\ \int h^{+}d\mu + \int f^{-} d\mu + \int g^{-}d\mu&= \int h^{-} d\mu + \int f^{+} d\mu + \int g^{+} d\mu \\ \int h^{+} d\mu - \int h^{-} d\mu&= \int f^{+} d\mu - \int f^{-} d\mu + \int g^{+} d\mu - \int g^{-} d\mu \\&= \int f d\mu + \int g d\mu\end{align*}

by Proposition 16.2.7, so $\int f + g d\mu = \int f d\mu + \int g d\mu$.

Now suppose that $f, g$ are $\complex$-valued 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 16.3.5. 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 16.3.6 (Dominated Convergence Theorem [Theorem 2.24, Fol99]). Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n}\subset \mathcal{L}^{1}(X)$, and $f:X \to \complex$ such that

$f_{n} \to f$ pointwise.
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. 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 16.3.4,

\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*}

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

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

Remark 16.3.7 (There is no dominated convergence theorem for nets). 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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16.3 Integration of Complex-Valued Functions

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