How can we bound $ch(G\cup H)$, the choice number, in terms of $ch(G)$ and $ch(H)$? It is a well-known proposition in graph theory that we can bound the chromatic number in the following way: $\chi(G\cup H)\leq \chi(G)\chi(H)$.  Here we could think of $G$ and $H$ as a $2$-edge coloring of $G\cup H$, so we have a product bound of $\chi$ along edge colorings or edge decompositions.  
Is there a similar bound for $ch(G)$, the choice number, also known as the list chromatic number of $G$?  My gut tells me that the bound should be a sum rather than a product, but I see $ch$ discussed in depth in very few textbooks and none appear to produce such a bound.  Even stronger, is it true that if $G$ is $g$-choosable and $H$ is $h$-choosable, then $G\cup H$ is $(g+h)$-choosable?
I should mention that I am hoping for a bound not dependent on any other graph invariants, but I'd be curious if others have thoughts or doubts or counterexamples.  
For a motivating example, we consider $\chi$ and $ch$ in the setting where $G = K_{n,n}$ and $H = 2K_n$ so that $G\cup H = K_{2n}$.  Clearly here we have $\chi(G\cup H) = ch(G\cup H)$.  Likewise, $\chi(H) = ch(H) = n$.  However, they disagree in that $\chi(G) = 2$ and $ch(G)>2$ (at least if $n>2$).  You can demonstrate fairly easily that $ch(K_{n,n}) \leq n$ by the following inductive procedure: select $u$ from the left and $v$ from the right so that $L(u)\neq L(v)$.  Then color $u$ and $v$ with different colors, delete $c(u)$ and $c(v)$ from everyone else's list, and apply the bound with $K_{n-1,n-1}$.  If No such pair $u,v$ exists, then it follows that $L(u) = L(v)$ for all $u,v$ and since then $|L(u)| = n$, we are in the case of a coloring all vertices using $n$ distinct coloring, but $K_{n,n}$ is bipartite.  
 So in this motivating example, $2n = \chi(G\cup H)\leq\chi(G)\chi(H) = 2\cdot n$ is tight and the guess $2n = ch(G\cup H)\leq ch(G)+ch(H) \leq n+n$ would be tight if we also knew that $ch(K_{n,n}) = n$.  However, $ch(K_{n,n})$ appears to be unknown in the literature.  
My motivating example is false, as pointed out in the comments. 
 A: Here are the fruits of further research.
It's an open problem whether $\operatorname{ch}(G \cup H) \le \operatorname{ch}(G) \cdot \operatorname{ch}(H)$, but it would follow from the $(a:b)$-choosability conjecture of Erdős, Rubin, and Taylor. To elaborate: an $(a:b)$-choosable graph is a graph in which, for any assignment of lists of size $a$ to the vertices, we can choose $b$ elements from each vertex's list so that the elements chosen for adjacent vertices are disjoint. A $k$-choosable graph is clearly $(k:1)$-choosable; the conjecture says that an $(a:b)$-choosable graph is also $(am:bm)$ choosable for any $m>1$.
Lemma. If $G$ is $(kl:l)$-choosable and $H$ is $l$-choosable, then $G \cup H$ is $kl$-choosable.
Proof. Given any assignment of lists of size $kl$ to the vertices of $G\cup H$, choose sublists of size $l$ such that, for vertices adjacent in $G$, their sublists are disjoint. Use these to list-color $H$. The colors given to vertices adjacent in $H$ must be different by definition; the colors given to vertices adjacent in $G$ are different because they came from disjoint sublists. Therefore we've list-colored $G \cup H$. $\square$
If the $(a:b)$-choosability conjecture holds, then knowing that $G$ is $k$-choosable would imply that it's $(kl:l)$-choosable, so this lemma would be a proof that $\operatorname{ch}(G \cup H) \le \operatorname{ch}(G) \cdot \operatorname{ch}(H)$.
Zuta and Voigt (1996) show that Every $2$-choosable graph is $(2m:m)$-choosable and therefore this list-coloring bound holds whenever either $G$ or $H$ has list chromatic number $2$.

On the other hand, according to Graph Coloring Problems by Jensen and Toft (which you might have access to from the publisher here), we can write down some function $f$ such that $\operatorname{ch}(G \cup H)$ is bounded by $f(\operatorname{ch}(G), \operatorname{ch}(H))$.
But I'm not sure why. Jensen and Toft cite the argument as [Alon, personal communication in 1994], and claim that it follows from Theorem 5.1 in Restricted colorings of graphs (Alon, 1993). This theorem 5.1 says that if a graph is $k$-choosable, then its average degree $d$ satisfies $$d \le 4 \binom{k^4}{k} \log \left(2 \binom{k^4}{k}\right).$$
I can't quite complete the argument from here. If $G$ is $k$-choosable and $H$ is $l$-choosable, we have upper bounds on the average degree of $G$ and $H$, and therefore an upper bound on the average degree of $G \cup H$. But it's certainly possible to have graphs with fixed average degree $d$ and arbitrarily large list chromatic number: just pick your favorite graph that has the list chromatic number you want, and tack on lots of low-degree vertices to bring the average degree down to $d$. So what do we do here?
