Let $1\leq p<\infty$. Suppose that $\{f_k\}$ is a sequence in $L^p(X,\mathcal{M},\mu)$ such that the limit
$f(x)=\lim_{k \to \infty}f_k(x)$
exists for $\mu$-a.e. $x\in X$. Asumme that
$\liminf_{k \to \infty} \|f_k \|_p=a$ is finite. Then prove that
a) $f \in L^P$
b) $\|f \|_p\leq a$
c) Assuming that $\|f \|_p=\lim_{k\to \infty}\|f_k \|_p$,
$\lim_{k\to \infty} \|f-f_k\|_p=0$
By Fatous' Lemma, I proved b). Is the fact that $f(x)=\lim_{k \to \infty}f_k(x)$ exists $\mu$-a.e. guarantees a)?.
For c), using inequality $(a+b)^p \leq 2^{p-1}(a^p+b^p)$, I hope
$|f-f_k|^p \leq 2^{p-1}(|f|^p+|f_k|^p)$ hold. Then I can use dominated convergence theorem. But with $f\in L^p$, I don't know how to proceed.
Any help will be thankful.
For c), apply Fatou's lemma to $2^{p-1}(|f|^p + |f_k|^p) - |f - f_k|^p$, giving
$$\liminf_{k \to \infty} \int_X 2^{p-1}(|f|^p + |f_k|^p - |f - f_k|^p)\, d\mu\ge 2^p \int_X |f|^p\, d\mu.$$
Thus
$$2^{p-1}(\|f\|_p^p + \liminf_{k\to \infty} \|f_k\|_p^p) - \limsup_{k\to \infty} \|f - f_k\|_p^p \ge 2^p\|f\|_p^p.$$
Since $\|f_k\|_p^p \to \|f\|_p^p$, we have
$$2^p\|f\|_p^p - \limsup_{k\to \infty} \|f - f_k\|_p^p \ge 2^p \|f\|_p^p.$$
As $\|f\|_p^p < \infty$, we deduce
$$0\ge \limsup_{k\to \infty} \|f - f_k\|_p^p .$$
Hence $\|f - f_k\|_p \to 0$.