Since $A \cup B = B$, it means that $A$ must be a subset of $B$, which is denoted by $A \subseteq B$. We'll explore two cases based on whether $A$ is empty or not.

1. First, consider the case where $A$ is empty, which we denote by $A = \emptyset$. This occurs if the condition for set $A$ doesn't hold for any real number, that is, if $m+1 > 3m-1$, which simplifies to $m < 1$.

2. Now consider the case where $A$ is non-empty ($A \neq \emptyset$). In this case, the inequalities defining set $A$ must align with the inequalities defining set $B$. This gives us the following system of inequalities:

$$\begin{cases} m+1 \leq 3m-1 \\ 1 \leq m+1 \\ 3m-1 \leq 10 \end{cases}$$

Solving this system, we have:
- From the first inequality: $m \geq \frac{2}{2} = 1$.
- From the second inequality: $m \geq 1-1 = 0$.
- From the third inequality: $3m \leq 11 \Rightarrow m \leq \frac{11}{3}$.

All three inequalities must be satisfied simultaneously, so combining them, we have $1 \leq m \leq \frac{11}{3}$.

Combining both cases, the range for $m$ must satisfy $m \leq \frac{11}{3}$. Therefore, the value range for $m$ can be described as $\boxed{\{m \mid m \leq \frac{11}{3}\}}$.