We assume that the CDF of some distribution $F_{X_k}(x)$ for $k=1,2,\cdots,K$. And that $X_k$ for all $k$ are mutually independent. Let us denote $X_{\max} = \max_{k=1,2\cdots,K}\{X_k\}$. Then the CDF of $X_{\max}$ is given by $F_{X_{\max}}(x) = \prod^{K}_{k=1}F_{X_k}(x)$.
$$\lim_{K\rightarrow \infty}\int^\infty_{-\infty}F_{X_{max}}(x)dx = \int^\infty_{-\infty}\lim_{K\rightarrow \infty}F_{X_{max}}(x)dx = 0$$
Can you prove the above equation?
Define $$f_K(x):=F-F^{K+1}_{X_\max}(x)$$ Since $F_{X_k}(x)\le 1$ for any $x\in \mathbb R$, you have that $$F^K_{X_\max}(x)=\prod_{k=1}^KF_{X_k}(x)\le \prod_{k=1}^{K+1}F_{X_k}(x)=F^{K+1}_{X_\max}(x)$$ Hence $0\le f_K(x)\le f_{K+1}(x)\le +\infty$. Moreover $$\lim_{K\to \infty}f_K(x)=0$$ for any $x\in\mathbb R. Hence, you may apply the Monotone Convergence Theorem to interchange the limit with the integral.