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

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.

$$P$$ 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.