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