Creating a bipartite graph with prescribed degrees Let $G=(U,V,E)$ denote a bipartite graph with vertices partitioned as $U \cup V$, $U\cap V=\emptyset$ and edges $E$ (so $E\subseteq U\times V$).  We also know that the degree sum equation must hold, namely, 
$$\sum_{u\in U}\deg(u) = \sum_{v \in V} \deg(v).$$
I was wondering given $U$ and $V$ and some prescribed degrees $d_u, u\in U$ and $d'_v, v\in V$ (that satisfy the degree sum equation, namely $\sum_{u} d_u = \sum_{v} d'_v$), can we find a bipartite graph $G=(U,V,E)$ that satisfies $\deg(u)=d_u$ and $\deg(v)=d'_v$ for all $u$ and $v$?

Thanks to all helpful comments, I understood that, if we allow multiple edges, we can always find the graph.  
Looking back into the application, I realize that I can allow multiple edges, so this is not a problem, however, there is one more constraint: In fact, we already have a simple (i.e., without multiple edges) bipartite graph $G'=(U,V,E')$ with degree sequences $(\hat{d}_u,~u\in U)$ and $(\hat{d}'_v, v \in V)$ that are all strictly bigger than the prescribed degrees, that is,
$$1 \le d_u \le \hat{d}_u-1, \qquad \forall u \in U$$ and $$1\le d'_v \le \hat{d}'_v-1, \qquad \forall v\in V.$$ 
The question is if we can find the graph $G=(U,V,E)$ with prescribed degrees $d_u$ and $d'_v$ without an edge $(u,v)$ if $(u,v)\not\in E'$ (i.e., without connecting the vertices that are not connected in $G'$ (But we can have multiple edges between two vertices if they are already connected in $G'$).

In general this is not true for simple graphs (i.e. there are no loops and no multiple edges).
Consider the example of $$ \sum_{u \in U} d_u=2+2+2+2+2=10=\sum_{v \in V} d_v.$$ So we want to construct a bipartite graph with only $1$ vertex in $V$ with degree $10$. But in $U$ we have exactly $5$ vertices. So we get $$10=deg(v) \le 5.$$ This yields a contradiction. 
We could generalize this construction and say that no $d_u$ ( resp. $d_v$) can be strictly bigger then $|V|$ (resp. $|U|$).
