Complexification has a canonical conjugation with fixed algebra g zero

The bracket formula of [L3] is $\mathbb{R}$-bilinear in the pair $(X\otimes z,Y\otimes w)$ and vanishes on each tensor-product relation in either slot, since $[X+X^{\mathrm{\prime }},Y]\otimes zw=[X,Y]\otimes zw+[X^{\mathrm{\prime }},Y]\otimes zw$ and $[cX,Y]\otimes zw=c[X,Y]\otimes zw$ for $c\in \mathbb{R}$; hence [L2] gives a well-defined $\mathbb{R}$-bilinear map ${\mathfrak{g}}_{\mathbb{C}}\times {\mathfrak{g}}_{\mathbb{C}}\to {\mathfrak{g}}_{\mathbb{C}}$ with the stated values on pure tensors. It is complex-bilinear because $[(tZ),W]=t[Z,W]$ and $[Z,tW]=t[Z,W]$, both checked on pure tensors and extended by $\mathbb{R}$-linearity in each argument.

The assignment $\theta$ is additive on pure tensors and vanishes on every relation of [L1], since $(X+X^{\mathrm{\prime }})\otimes \bar{z}=X\otimes \bar{z}+X^{\mathrm{\prime }}\otimes \bar{z}$, $X\otimes \overline{z+z^{\mathrm{\prime }}}=X\otimes \bar{z}+X\otimes \overline{z^{\mathrm{\prime }}}$, and conjugation is $\mathbb{R}$-linear, so that the relation $(X,cz)-(cX,z)$ for $c\in \mathbb{R}$ becomes $X\otimes c\bar{z}-cX\otimes \bar{z}=0$; by [L1] and the additive generation of ${\mathfrak{g}}_{\mathbb{C}}$ by elementary tensors, $\theta$ extends uniquely to a well-defined additive map $\sigma \text{\hspace{0.17em}⁣}:{\mathfrak{g}}_{\mathbb{C}}\to {\mathfrak{g}}_{\mathbb{C}}$ with $\sigma (X\otimes z)=X\otimes \bar{z}$.

For pure tensors $u=X\otimes z$, $v=Y\otimes w$, alternation in ${\mathfrak{g}}_0$ implies $[X,Y]=-[Y,X]$ by expansion of $[X+Y,X+Y]=0$. Thus $[u,v]+[v,u]=([X,Y]+[Y,X])\otimes zw=0$ and $[u,u]=0$. For an arbitrary finite tensor sum $Z={\sum }_a{u}_a$, bilinearity now gives $[Z,Z]={\sum }_a[u_a,u_a]+{\sum }_{a<b}([u_a,u_b]+[u_b,u_a])=0$. Jacobi, unlike alternation, is trilinear: on three pure tensors it is the Jacobi expression in ${\mathfrak{g}}_0$ tensored with the product of their scalars, hence zero, and trilinearity extends it to arbitrary finite sums. This proves the complex Lie-algebra axioms.

$\sigma$ is conjugate-linear: it is additive by construction, and $\sigma (t(X\otimes z))=X\otimes \overline{tz}=\bar{t}(X\otimes \bar{z})=\bar{t}\sigma (X\otimes z)$ for complex $t$; both sides are additive in the argument, so the identity extends to all of ${\mathfrak{g}}_{\mathbb{C}}$. It is involutive because ${\sigma }^2(X\otimes z)=X\otimes z$ on pure tensors and ${\sigma }^2$ is additive. Finally $\sigma$ preserves brackets: on pure tensors $\sigma ([X\otimes z,Y\otimes w])=[X,Y]\otimes \overline{zw}=[X\otimes \bar{z},Y\otimes \bar{w}]=[\sigma (X\otimes z),\sigma (Y\otimes w)]$, and since both sides are additive and the bracket and $\sigma$ are additive, the identity extends to all pairs.

Choose a real basis $e_1,\dots ,e_n$ of ${\mathfrak{g}}_0$. Every tensor has an expression $Z={\sum }_j{e}_j\otimes z_j$: expand each real factor of a finite sum of pure tensors in this basis and use the tensor relations. This expression is unique, because for the dual basis functional $e_j^{\ast }$ the bilinear map $(Y,z)\mapsto e_j^{\ast }(Y)z$ induces by [L2] a real-linear map ${\phi }_j:{\mathfrak{g}}_{\mathbb{C}}\to \mathbb{C}$ with ${\phi }_j(Z)=z_j$. Consequently $\sigma Z=Z$ if and only if $\overline{z_j}=z_j$ for every $j$, equivalently $z_j\in \mathbb{R}$ for every $j$. In that case $Z={\sum }_j{e}_j\otimes z_j=({\sum }_j{z}_j{e}_j)\otimes 1=\epsilon ({\sum }_j{z}_j{e}_j)$, while every element of $\epsilon ({\mathfrak{g}}_0)$ is visibly fixed. Thus the fixed locus is exactly $\epsilon ({\mathfrak{g}}_0)$.

$\epsilon$ is injective: if $X\mathrm{\ne }0$, extend $X$ to an $\mathbb{R}$-basis of ${\mathfrak{g}}_0$ and let $\lambda$ be the dual basis functional with $\lambda (X)=1$; the $\mathbb{R}$-bilinear map $(Y,z)\mapsto \lambda (Y)z$ induces by [L2] an $\mathbb{R}$-linear $\phi \text{\hspace{0.17em}⁣}:{\mathfrak{g}}_{\mathbb{C}}\to \mathbb{C}$ with $\phi (Y\otimes z)=\lambda (Y)z$, so $\phi (X\otimes 1)=1$ and $X\otimes 1\mathrm{\ne }0$. Moreover $\epsilon ([X,Y])=[X,Y]\otimes 1=[X\otimes 1,Y\otimes 1]=[\epsilon X,\epsilon Y]$ by the bracket formula, so $\epsilon$ is an injective real Lie-algebra homomorphism. Together with step 3.1 this identifies ${\mathfrak{g}}_{\mathbb{C}}^{\sigma }$ with ${\mathfrak{g}}_0$ as a real Lie subalgebra.