Exercise 2.6.19 of Jech's Axiom of choice asks to show that the Hahn Banach theorem is equivalent to the existence of a real valued measure for all Boolean algebras over $ZF$. This is a sketch at my attempt:
Assuming the Hahn Banach theorem holds. And let $B$ be a boolean algebra, without choice we have that there is some set $S$ and a subalgebra $P\subseteq \mathcal{P}(S)$ such that there is a surjective homomorphism $h:P\rightarrow B$. Consider the space $\ell^\infty(S)$ then I have that $g:\mathcal{P}(S)\rightarrow \ell^\infty(S)$ where each set is sent to its characteristic function. We can consider the subspace generated by $E=\text{span}(g(P))$ and it's subspace $H=\text{span}(g(\text{ker}(h)))$ Then we have a space $E/H$ which is a vector space and there is an inclusion of $B$ in $E/H$. Consider the map that sends $1+H$ (the quotient class of the constant 1) to $1\in \mathbb{R}$ then by Hahn Banach it is extended to all of $E/H$. We need to assure that the function it extends to is a measure so we need the image of $B$ to be sent to $[0,1]$ to this end I define $p(f)$ to be $0$ if $f\in -\text{span}(B)$ and $p(f)=\|f\|_{\infty,E/H}$ otherwise. This function is subadditive and any extension that is dominated by $p$ will induce a finitely additive measure on $B$.
As for the other direction we have that over $ZF$ the Hahn Banach theorem is equivalent to showing every Banach space has a non $0$ functional. I believe every Banach space can be embedded in some $L^\infty(X,\mathcal{A},\mu)$. The dual of $L^\infty(X,\mathcal{A},\mu)$ is the set of finitely additive signed measures on $\mathcal{A}$ that are absolutely continuous with respect to $\mu$ but by assumption $\mathcal{A}/null(\mu)$ is a Boolean algebra and it will have a finitely additive measure which is non trivial.
Is this correct?
Edit: The second part is not correct since the embedding into $L^\infty $ requires Hahn Banach from what I recall
Edit: I actually managed to find a proof out there which is different than mine. I will add it to this question for completeness.
Let $B$ be a boolean algebra and consider the meet semilattice of the partitions of $B$ where $Q\leq P$ if $Q$ refines $P$. Observe that a function $f:P\rightarrow \mathbb{R}$ can be extended to $f_Q:Q\rightarrow \mathbb{R}$ naturaly by setting $f_Q(q)=f(p)$ where $p\in P$ is the unique element such that $q\leq p$. We define $S(B)=\{f:P\rightarrow \mathbb{R}: P \text{ is a partition of } B\}/\sim$ where $\sim$ is the relation where given $f:P\rightarrow \mathbb{R}$ and $g:Q\rightarrow \mathbb{R}$ we set $f\sim g$ iff $f_{Q\wedge P}=g_{Q\wedge P}$. This has a natural structure as a vector space. We can endow $S(B)$ with the upper limit norm or $\|f\|=\sup_{p\in P}f(p)$ which is easy to verify is well defined on the equivalence classes. We can identify $b\in B$ to the function $f:\{b,b^c\}\rightarrow \mathbb{R}$ where $f(b)=1$ and $f(b^c)=0$. We see that a finitely additive measure $m$ on $A\subseteq B$ will define on $E=\{f\in S(B): \text{dom}(f)\subseteq A \}$ (to be more precise the classes that have a representative with domain contained in $A$) we define $\varphi:E\rightarrow \mathbb{R}$ to be the function $\varphi(f)=\sum_{p\in \text{dom}(f)}f(p)m(p)$ since $m$ is finitely additive it is easy to verify that its well defined on the equivalence classes. We can extend such a function to all of $S(B)$ by Hahn Banach so that it is still of norm $1$ and it will define on $B\subseteq S(B)$ a finitely additive measure.
Source: "EL TEOREMA DE HAHN-BANACH COMO PRINCIPIO DE ELECCIÓN" by Xavier Caicedo & Germán Enciso