= Solution
A <simple function> is a measurable function with finitely many values, written $s=\sum_{j=1}^m a_j\mathbf1_{A_j}$ using disjoint measurable sets. For nonnegative $s$ its <Lebesgue integral> is $\mu(s)=\sum_ja_j\mu(A_j)$, 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 $\mu(f)=\sup\{\mu(s):s\text{ simple},\ 0\leq s\leq f\}$. 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 $0\leq f_n\uparrow f$ 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 $c\geq0$.
For integrable signed $f,g$, $|f+g|\leq|f|+|g|$ 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 $c\geq0$, 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>. \b[The indicated map is linear.]
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