Let $\Omega $ open bounded with Lipschitz boundary. Let $$m=\inf\left\{\int_\Omega f(x,u(x),\nabla u(x))\mathrm d x\mid u\in u_0+W_0^{1,p}(\Omega ) \right\},$$ and $I(u_0)<\infty $. I recall that $u\in u_0+W_0^{1,p}(\Omega )$ means that $u|_{\partial \Omega }=u_0.$
We suppose that there is $\alpha _1>0$ and $\alpha _3\in \mathbb R$ s.t. $$f(x,u,\xi)\geq \alpha _1|\xi|^p+\alpha _3.$$
Why this implies that $-\infty <m\leq I(u_0)<\infty $ ? The fact that $m\leq I(u_0)$ is fine because $u_0\in u_0+W_0^{1,p}(\Omega )$ and $I(u_0)<\infty $. Now, why the fact that $f(x,u,\xi)\geq \alpha _1|\xi|^p+\alpha _3$ give us that $m>-\infty $ ? I that this condition implies that $I(u)\geq \alpha _1\|\nabla u\|_{L^p}^p+\alpha _3m(\Omega )$ whenever $u\in u_0+W_0^{1,p}(\Omega )$ and thus that $$m\geq \inf\{\alpha _1\|\nabla u\|_{L^p}^p+\alpha _3\mid u\in u_0+W_0^{1,p}(\Omega )\}$$ but why $\inf\{\alpha _1\|\nabla u\|_{L^p}^p+\alpha _3\mid u\in u_0+W_0^{1,p}(\Omega )\}>-\infty $ ? In my course, it's written that $m>-\alpha _3m(\Omega )$ but I don't really see why.
I denote $\mu(\Omega )$ instead of $m(\Omega )$ for the measure of $\Omega $. As you said, $$I(u)\geq \alpha _1\|\nabla u\|_{L^p}^p+\alpha _3\mu(\Omega )\geq \alpha _3\mu(\Omega ), $$ and thus $$m\geq \alpha _3\mu(\Omega )>-\infty.$$
I think that your last inequality should be $m>-|\alpha _3|\mu(\Omega )$, but it's not necessary here. Notice that since $\alpha _3\in \mathbb R$, you have that $$\alpha _3=\text{sgn}(\alpha _3)|\alpha _3|\geq-|\alpha _3|,$$ is always true.