Consider an extension of algebraic number fields $L/K$ and their respective rings of integers $\mathcal{O}_L$, $\mathcal{O}_K$. They are defined as the integral closures of $\mathbb{Z}$ in their respective fields.
Is it the case that $\mathcal{O}_L$ is the integral closure of $\mathcal{O}_K$ in $L$?
Yes. Let $A$ be the integral closure of $\mathcal{O}_K$, if $x\in \mathcal{O}_L$, then $x$ is integral over $\mathbb{Z}$ so it's integral over $\mathcal{O}_K\supset \mathbb{Z}$, therefore $x \in A$. If $x\in A$ then by definition $x \in L$ and $x$ is integral over $\mathcal{O}_K$, because $\mathcal{O}_K$ is integral over $\mathbb{Z}$, $x$ is integral over $\mathbb{Z}$, therefore $x \in \mathcal{O}_L$. For this you have to know that the property of being integral is transitive. This statement can be found in Neukirch's Algebraic Number Theory, Theorem 2.4.