The Lebesgue integral is linear on $L^1(\mu )$

First treat real-valued $f,g$ and put $h:=f+g$. By [L4], the function $h$ is measurable. Also $h^{+}\le f^{+}+g^{+},h^{-}\le f^{-}+g^{-},$ so [L2] and [L3] give $\int h^{+}d\mu \le \int f^{+}d\mu +\int g^{+}d\mu <+\mathrm{\infty },\int h^{-}d\mu \le \int f^{-}d\mu +\int g^{-}d\mu <+\mathrm{\infty }\mathrm{.}$ Hence $h$ is integrable. Since $h^{+}+f^{-}+g^{-}=h^{-}+f^{+}+g^{+},$ another application of [L2] yields $\int h^{+}d\mu +\int f^{-}d\mu +\int g^{-}d\mu =\int h^{-}d\mu +\int f^{+}d\mu +\int g^{+}d\mu ,$ which rearranges to $\int (f+g)d\mu =\int fd\mu +\int gd\mu \mathrm{.}$

Now let $c\in \mathbb{R}$ and let $f$ be real-valued. By [L4], $cf$ is measurable. If $c\ge 0$, then $(cf{)}^{+}=c{f}^{+},(cf{)}^{-}=c{f}^{-};$ if $c<0$, then $(cf{)}^{+}=(-c)f^{-},(cf{)}^{-}=(-c)f^{+}\mathrm{.}$ In both cases [L3] shows that $cf$ is integrable and that $\int (cf)d\mu =c\int fd\mu \mathrm{.}$

Let $h=u+iv\in L^1(\mu )$ and $\gamma =a+ib\in \mathbb{C}$. Then $\gamma h=(au-bv)+i(av+bu)\mathrm{.}$ The real-valued functions $au-bv$ and $av+bu$ are integrable by steps 1.1 and 1.2, and their real-linear integral formulas combine into $\int (\gamma h)d\mu =\gamma \int hd\mu \mathrm{.}$ Writing $\alpha f=h_1+i{k}_1$ and $\beta g=h_2+i{k}_2$ with real-valued integrable $h_j,k_j$, step 1.1 gives $\int (\alpha f+\beta g)d\mu =\int (h_1+h_2)d\mu +i\int (k_1+k_2)d\mu =\int h_{1}d\mu +i\int k_{1}d\mu +\int h_{2}d\mu +i\int k_{2}d\mu =\alpha \int fd\mu +\beta \int gd\mu \mathrm{.}$