Convergence in Integral and Bounded in Lp Implies Convergence in Measure?

Question

I am attempting this problem:

Let $P \subset [1, \infty]$ be the set of all $p \in [1, \infty]$ for which the following statement is true:

If $f_n, f: [0, 1] \to > \mathbb{R}$ are integrable function such that $\sup_{n \in \mathbb{N}} > \| f_n \|_{L^p} < \infty$ and $\lim_{n \to \infty} \int_A f_n d\lambda = \int_A f d\lambda$ for every $A \in \mathcal{B}([0, 1])$, then $(f_n) \to f$ in measure.

Find $P$ and provide a counter example for all $p \in P^c \cap [1,\infty]$.

What I have tried so far seems to have failed. I notice that the measure space that we are dealing with is a finite measure space and thus the Egorov's Theorem holds. This means if we can show the function is almost everywhere pointwise convergent, then we are done. I have failed to do so.

Starting off with $p = \infty$, thinking it might be the easier case. We have a uniform bound on $f_n$ for all $n$ which implies that $f_n$ is integrable on $[0, 1]$. We have moreover that $\lim_{n \to \infty} \int_A f_n d\lambda = \int_A f d\lambda$ for all $A \in \mathcal{B}([0, 1])$, but then I will need some kind of partial converse of Lebesgue dominated convergence theorem to show almost pointwise convergence. But this is most likely not true. This probably implies that a counterexample exists, but I couldn't find one so far.

Answer 1

$P$ is empty! $f_n(x)=e^{2\pi in x}$ is bounded in $L^{p}$ for every $p \in [1,\infty]$ and $\int_Af_n(x)dx \to 0=\int_A 0\, dx$ for every $A$ by Riemann Lebesgue Lemma But $\lambda \{x: |f_n(x)-0|>\epsilon\}=1$ for every $\epsilon \in (0,1)$ so $f_n$ does not tend to $0$ in measure.

Ref: https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma