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