Maybe this is a silly question but this has been bothering me for some time. Let $\Phi$ be a root system with a base $\Delta=\{\alpha_1,\ldots,\alpha_l\}$. Then $\Delta^\vee=\{\alpha_1^\vee,\ldots,\alpha_l^\vee\}$ is also a base, where $\alpha_j^\vee=\frac{2\alpha_j}{(\alpha_j,\alpha_j)}$ for each $j=1,\ldots,l$. Let $\lambda_1,\ldots,\lambda_l$ be the basis dual to $\Delta^\vee$ with respect to the inner product, i.e. $(\lambda_i,\alpha_j^\vee)=\delta_{ij}$. Humphreys, in Section 13 of his text on Lie algebras, then states that the $\lambda_i$ are dominant weights and he says we call them the fundamental dominant weights. However, I do not see why they are weights.
Humphreys defines a weight as an element $\lambda\in E$ such that $\langle\lambda,\alpha\rangle:=\frac{2(\lambda,\alpha)}{(\alpha,\alpha)}\in\mathbb{Z}~\forall~\alpha\in\Phi$, where $E$ is the underlying Euclidean space. However, the way that we have defined the $\lambda_i$ only shows that $\langle\lambda,\alpha\rangle\in\mathbb{Z}~\forall~\alpha\in\Delta$. As $\langle\cdot,\cdot\rangle$ is not linear in the second component I do not see how this extends to every root in $\Phi$.
It boils down to the fact that if $\Delta$ is a base of $\Phi$, then $\Delta^\vee=\{\alpha_1^\vee,\dots,\alpha_l^\vee\}$ is a base of the dual root system $\Phi^\vee$. You can find a proof on this site here, beware however that there are a few minor typos in the accepted answer, but the idea is correct.
Suppose $\lambda$ is such that $\langle\lambda,\alpha_i\rangle\in\mathbb{Z}$ for all $\alpha_i\in\Delta$. If $\alpha\in\Phi$ is an arbitrary root, we can write $\alpha^\vee=\sum_i c_i\alpha_i^\vee$ for some $c_i\in\mathbb{Z}$. Then since the form $(-,-)$ is bilinear, $$ \langle\lambda,\alpha\rangle=(\lambda,\alpha^\vee)=(\lambda,\sum_i c_i\alpha_i^\vee)=\sum_i c_i(\lambda,\alpha_i^\vee)=\sum_i c_i\langle\lambda,\alpha_i\rangle\in\mathbb{Z}, $$ since the $c_i$ and $\langle\lambda,\alpha_i\rangle$ are all integers. Hence $\lambda$ is a weight.