Does any complex representation of $\mathfrak{g}_\mathbb{C}$ define a real representation of $\mathfrak{g}$? Proposition 3.39 of Hall's Lie Groups, Lie Algebras And Representations:
"Let $\mathfrak{g}$ be a real Lie algebra, $\mathfrak{g}_\mathbb{C}$ its complexification, and $\mathfrak{h}$ an
arbitrary complex Lie algebra. Then every real Lie algebra homomorphism of $\mathfrak{g}$ into $\mathfrak{h}$ extends uniquely to a complex Lie algebra homomorphism of $\mathfrak{g}_\mathbb{C}$ into $\mathfrak{h}$."
In particular this means that any real representation of $\mathfrak{g}$ defines a complex representation of $\mathfrak{g}_\mathbb{C}$.
Question: Does the converse hold? Does any complex representation of $\mathfrak{g}_\mathbb{C}$ define a real representation of $\mathfrak{g}$? Are there any conditions for when this may or may not hold?
 A: Of course it does. If the Lie algebra $\mathfrak{g}_{\mathbb C}$ acts on a complex vector space $V$, simply consider the restriction of this action to $\mathfrak g$. To be more precise, if $X\in\mathfrak g$ and if $v\in V$, then define $X.v$ as $(X\otimes 1).v$. I am assuming here that Hall defined $\mathfrak{g}_{\mathbb C}$ as $\mathfrak{g}\bigotimes_{\mathbb{R}}\mathbb{C}$. If he used another definition, please say which one did he use.
A: The comment section to the other answer brought up an issue which maybe deserves another answer.
What is true:
For a real Lie algebra $\mathfrak g$, there is a one-to-one correspondence between

*

*representations of $\mathfrak g$ on complex vector spaces; that is, $\mathbb R$-linear Lie algebra homomorphisms $\mathfrak g \rightarrow \mathrm{End}_{\mathbb C}(V)$ for a complex vector space $V$, and

*representations of the complexification $\mathfrak g_\mathbb C := \mathbb C \otimes_{\mathbb R} \mathfrak g$ on complex vector spaces; that is, $\mathbb C$-linear Lie algebra homomorphisms $\mathfrak g_\mathbb C \rightarrow \mathrm{End}_{\mathbb C}(V)$ for a complex vector space $V$,

Namely, the quote from Hall's book in the OP explicitly states that $\mathfrak h$ is a complex Lie algebra, and Hall and José Carlos Santos' answer apply this to the case $\mathfrak h = \mathrm{End}_{\mathbb C}(V)$ for a complex vector space $V$.
This equivalence is kind of common knowledge and comes up regularly on this site. Cf. In what sense are complex representations of a real Lie algebra and complex representations of the complexified Lie algebra equivalent?, https://math.stackexchange.com/a/3258221/96384, Bijection between the complex representations of a real Lie algebra and the complex representations of its complexification, Obtaining representation of a real Lie algebra from the complexification "by restriction".
Where one has to be careful:
If in the title and the question body the phrase "real representation of $\mathfrak g$" is supposed to mean a representation of $\mathfrak g$ on a real vector space $W$, i.e. a Lie algebra homomorphism $\mathfrak g \rightarrow \mathrm{End}_{\mathbb R}(W)$, then

*

*such a representation also does define a complex representation of $\mathfrak g_\mathbb C$, but that needs an extra step absent from the construction in the quote from Hall;

*it is not true that this process is "reversible" in general; indeed, not only can it happen that non-isomorphic real representations of $\mathfrak g$ give out isomorphic complex representations of $\mathfrak g_\mathbb C$ (as John Baez pointed out), but more strikingly, not every complex representation of $\mathfrak g_{\mathbb C}$ comes from a real representation of $\mathfrak g$ via such procedure.

Ad 1: The extra step is that we first have to complexify the real vector space $W$ to $W_\mathbb C := W \otimes_\mathbb R \mathbb C$, and extend our real representation $\mathfrak g \rightarrow \mathrm{End}_{\mathbb R}(W)$ via the natural $\mathbb R$-linear inclusion $\mathrm{End}_{\mathbb R}(W) \hookrightarrow \mathrm{End}_{\mathbb C}(W_\mathbb C)$. Only then do we have a map $\mathfrak g \rightarrow \mathrm{End}_{\mathbb C}(V)$ (namely, $V := W_\mathbb C$) as in what was true above, and can proceed like there.
Ad 2: But it is precisely this extra step which is not reversible. Note also that in José Carlos Santos' answer, what comes out is a complex representation of $\mathfrak g$, i.e. a map $\mathfrak g \rightarrow \mathrm{End}_{\mathbb C}(V)$ with a complex vector space $V$. There is no reason, and in general it is not true, that that complex vector space $V$ has a real subspace $W$ which is stable under the Lie algebra action and whose complex span is all of $W$.
As a standard example, consider the obvious complex representation of the real Lie algebra $\mathfrak{su}_2$ on $V:=\mathbb C^2$ via writing $\mathfrak{su}_2$ as the matrices $\pmatrix{ai & b+ci\\-b+ci &-ai}$ with $a,b,c \in \mathbb R$. If there were a two-(real-)dimensional subspace $W$ of $V$ stable under the $\mathfrak{su}_2$-action, we would have a nonzero map $\mathfrak{su}_2 \rightarrow \mathfrak{gl}_\mathbb R(W) \simeq \mathfrak{gl}_2(\mathbb R)$. It's easily seen that such a map would induce an isomorphism $\mathfrak{su}_2 \simeq \mathfrak{sl}_2(\mathbb R)$, which is absurd for a plethora and more of reasons.
It is, rather, an interesting question to decide for a given real Lie algebra $\mathfrak g$, which of its complex representations "come" from real representations in such way, which ones are "truly complex", and as it turns out there is a third case, namely they can come from "quaternionic" representations. Cf. part A of https://math.stackexchange.com/a/4026224/96384 as well as What property of the root system means a Lie algebra has complex structure?, or the discussion in https://math.stackexchange.com/a/3712110/96384.
