 Solution

A simple function is a measurable function with finitely many values, written $s={\sum }_{j=1}^m{a}_j{1}_{A_j}$ using disjoint measurable sets. For nonnegative $s$ its Lebesgue integral is $\mu (s)={\sum }_j{a}_j\mu (A_j)$, with $0\cdot \infty =0$. An integrable signed simple function has ${\sum }_j|a_j|\mu (A_j)<\infty$ and the same signed sum defines its integral.

For a nonnegative measurable $f$ define $\mu (f)=\sup \{\mu (s):s\text{ simple},0\le s\le f\}$. A real measurable $f$ is Lebesgue integrable when $\mu (|f|)<\infty$; then $\mu (f)=\mu (f^{+})-\mu (f^{-})$, both terms finite. The monotone convergence theorem states that if $0\le f_n\uparrow f$ pointwise, or almost everywhere, then $\mu (f_n)\uparrow \mu (f)$, allowing infinity.

For nonnegative $f,g$, choose nonnegative simple approximations $s_n\uparrow f$, $t_n\uparrow g$. Their sums increase to $f+g$; simple-function additivity and monotone convergence give $\mu (f+g)=\mu (f)+\mu (g)$. The same approximation proves $\mu (cf)=c\mu (f)$ for $c\ge 0$.

For integrable signed $f,g$, $|f+g|\le |f|+|g|$ makes their sum integrable. Write $u=f^{+}+g^{+}$, $v=f^{-}+g^{-}$. The identity $u+(f+g{)}^{-}=v+(f+g{)}^{+}$ and nonnegative additivity, with finite integrals, give $\mu (f+g)=\mu (f)+\mu (g)$. Also $(cf{)}^{\pm }=c{f}^{\pm }$ when $c\ge 0$, while negative $c$ interchanges the two parts, so $\mu (cf)=c\mu (f)$ for every real $c$. Integrals do not change upon alteration on a null set, hence this descends to Lebesgue space. The indicated map is linear.