Sequence of functions convergent almost everywhere and bounded in $L^p$-norm implies convergence in $L^p$

Question

Question:

Suppose that $f_n \in L^1(I=(-1,1),\lambda^1)$ such that $f_n \to f$ as $n \to \infty$ $\lambda^1$-a.e.. Also, suppose there exists a finite constant $M>0$ such that

$$\int_I |f_n|^{1+\delta} d\lambda^1 \leq M~~~~j=1,2,\dots$$

with some fixed $\delta \geq 0$. Does it follow that $\|f_n - f\|_{L^1} \to 0$ if $\delta=0$? If $\delta>0$?

Thoughts:

I know that $f_n \to f$ as $n \to \infty$ $\lambda^1$-a.e. implies convergence in $\lambda^1$ and I wanted to try to potentially apply that. I was also thinking that I could try to apply Dominated Convergence Theorem using the fact that the integrals are bounded... but the sequence being bounded in $L^p$ doesn't necessarily mean the sequence itself is bounded.

Also, I am thinking that we don't have convergence in $L^1$ when $\delta=0$, but we do when $\delta>0$.

Answer 1

As $\mu$ is finite, being bounded in $L^p$, for some $p>1$, implies being uniformly integrable and then, by the Vitali convergence theorem, we can conclude the convergence in $L^1$. I recommend you read about Vitali's theorem because a direct approach in this problem is almost the same as to prove this (very important) theorem.

Also, being bounded in $L^1$ is not enough to conclude converge.