10. If the set $A=\{x \mid-2 \leqslant x \leqslant 5\}, B=\{x \mid m+1 \leqslant$ $x \leqslant 2 m-1\}$, and $A \cap B=B$, then the range of real number $m$ is $\qquad$.
$$
\text { 10. } m \leqslant 3 \text {. }
$$

From $A \cap B=B$, we know that $B$ is a subset of $A$.
When $B=\varnothing$, we have $m+1>2 m-1$, which gives $m<2$;
When $B \neq \varnothing$, we have
$$
\left\{\begin{array}{l}
-2 \leqslant m+1, \\
2 m-1 \leqslant 5, \\
m+1 \leqslant 2 m-1 .
\end{array}\right.
$$

Solving this, we get $2 \leqslant m \leqslant 3$. Therefore, $m \leqslant 3$.