In the book Dacorogna - "Direct methods in the calculus of variations", corollary 3.22 on page 94 states:
Let $p \geq 1$, $\Omega \subseteq \mathbb R^n$ open and $f:\Omega \times \mathbb R^{N \times n} \to \mathbb R$ be a continuous function satisfying
$$f(x,\xi) \geq \langle a(x), \xi \rangle + b(x)$$
for some $a \in L^{p'}$, $1/p+1/p'=1$, $b \in L^1$. Let
$$I(u)=\int_\Omega f(x, \nabla u(x)) dx.$$
Then $I(u)$ is lower semicontinuous.
Proof: define $g(x,\xi)=f(x,\xi)-\langle a(x), \xi \rangle + b(x)$. If $$G(u)=\int_\Omega g(x, \nabla u(x)) dx$$ is lsc then $$I(u)=\int_\Omega f(x, \nabla u(x)) dx$$ is also lsc, hence we can assume $f \geq 0$ without loss of generality.
Let us take $\xi_\nu \overset{L^p}{\to}\xi$. There exist a subsequence which converges almost everywhere $\xi_{\nu_k}(x) \to \xi(x)$ hence we can use Fatou's lemma
$$\liminf_{\nu_k} \int_\Omega f(x,\xi_{\nu_k}(x))dx \geq \int_\Omega \liminf_{\nu_k} f(x,\xi_{\nu_k}(x))dx$$ and the proof ends with the statement "Combining the above inequality with the lower semicontinuity of f, we have the claim."
My question is: isn't the inequality only valid for a particular subsequence? How do you conclude that it is valid also for $\xi_\nu$?
The term \begin{align*} \liminf_{\nu}\int_{\Omega}f(x,\xi_{\nu}(x))dx \end{align*} can be realized to a subsequence \begin{align*} \lim_{\nu_{k}}\int_{\Omega}f(x,\xi_{\nu_{k}}(x))dx. \end{align*} Meanwhile the subsequence $\{\xi_{\nu_{k}}\}$ has a further subsequence $\{\xi_{v_{k_{l}}}\}$ such that $\xi_{\nu_{k_{l}}}(x)\rightarrow\xi(x)$ a.e.
Following the reasoning in the reference, we get \begin{align*} \liminf_{\nu_{k_{l}}}\int_{\Omega}f(x,\xi_{\nu_{k_{l}}}(x))dx\geq I(\xi). \end{align*} But then \begin{align*} \liminf_{\nu_{k_{l}}}\int_{\Omega}f(x,\xi_{\nu_{k_{l}}}(x))dx=\lim_{\nu_{k}}\int_{\Omega}f(x,\xi_{\nu_{k}}(x))dx \end{align*} since the former sequence is a subsequence of the later convergent sequence.