Existence of variation Let $I[w] =\int_U L(Dw,w,x) dx$. Let $1&lt;q&lt;\infty$, and there exist constants $\alpha&gt;0$,$\beta\ge0$ such that $$L(p,z,x)\ge \alpha |p|^q - \beta$$

This implies that if $I[w]$ exists,
$$I[w] \ge \alpha \|Dw\|_{L^q}^q -\beta |U|$$

Now Evans says that for $w\in W^{1,q}(U)$ that $I[w]$ is defined but possibly infinite. What is the reason for this? Is it true always that functions which are bounded below are integrable (where integral can be $\infty$)?
On a set of finite measure (such as $U$ here) every measurable function that is bounded from below has a well-defined integral, which is either a real number or $+\infty$. Indeed, one can write $f=f^+-f^-$ where $f^+=\max(f,0)$ and $f^-=\max(-f,0)$. Then define $\int_U f=\int_U f^+-\int_U f^-$.  The second integral is finite, the first may be $+\infty$.

In this case, $L(Dw(x),w(x),x)$ is bounded from below by $-\beta$. One still has to check that it's measurable, which requires placing some condition on $L$ (continuity is enough, but there are more general conditions that suffice for this purpose).
On a set of finite measure (such as $U$ here) every measurable function that is bounded from below has a well-defined integral, which is either a real number or $+\infty$. Indeed, one can write $f=f^+-f^-$ where $f^+=\max(f,0)$ and $f^-=\max(-f,0)$. Then define $\int_U f=\int_U f^+-\int_U f^-$.  The second integral is finite, the first may be $+\infty$.

In this case, $L(Dw(x),w(x),x)$ is bounded from below by $-\beta$. One still has to check that it's measurable, which requires placing some condition on $L$ (continuity is enough, but there are more general conditions that suffice for this purpose).