Let $(E,\mathcal{A},\mu)$ be a measure space. Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of measurable functions, which converges in measure to $f:\forall \epsilon>0,\lim_n\mu(\left\{|f_n-f|>\epsilon \right\})=0.$
- Suppose that $\mu(E)<+\infty.$ Prove that $\forall k \in \mathbb{N},\lim_n\int_E \min(|f_n|,k)d\mu=\int_E \min(|f|,k)d\mu.$
Deduce that $$\int_E |f|d\mu\leq \liminf_n\int_E |f_n|d\mu.$$
- We don't suppose that $\mu(E)<+\infty$ (it can be infinite). Prove that :$$\int_E|f| d\mu\leq \liminf_n \int_E|f_n|d\mu.$$
I solved 1. : for $ \epsilon>0$, for $k \in \mathbb{N},\forall n\in \mathbb{N}$,$$\int_E|\min(|f_n|,k)-\min(|f|,k)|d\mu \leq\epsilon \mu(E)+2k\mu(\left\{|\min(|f_n|,k)- \min(|f|,k)|>\epsilon\right\})$$ so we have $$\limsup_n\int_E|\min(|f_n|,k)-\min(|f|,k)|d\mu \leq \epsilon\mu(E)$$ since $\min(|f_n|,k)$ converges in measure to $\min(|f|,k)$ ($\mu(E)<\infty$), $\epsilon$ is arbitrary, we deduce that $\forall k \in \mathbb{N},\lim_n\int_E \min(|f_n|,k)d\mu=\int_E \min(|f|,k)d\mu.$
- Deduction : $\int_E \min(|f|,k)d\mu=\liminf_n\int_E\min(|f_n|,k)d\mu \leq\liminf_n\int_E|f_n|d\mu.$ We conclude by the monotone convergence theorem.
For 2. the professor told us to find a way similar to 1. but the problem is that $\mu(E)$ can be infinite.
So is there a way similar to 1. to solve 2.?