Let $(X,\Sigma,\mu)$ a measure space and $f$ a measurable function and $A \in \Sigma$.
Also let $\{E_n:n \in \Bbb{N}\}\subseteq \Sigma$ and $E_n \subseteq A,\forall n \in \Bbb{N}$ and assume that $1_{E_n} \to f$ in the mean,or in measure or almost uniformly on $A$.Prove then that $\exists E \subseteq A$ such that $f=1_{E}$ on $A$
Proof:
We know that convergence in measure and in the mean,imply that exists a subsequence $1_{E_{n_k}} \to f$ pointwise a.e
But thus $\limsup_k 1_{E_{n_k}}=f$ a.e on $A$
Also $\limsup_k 1_{E_{n_k}}=1_{\limsup_kE_{n_k}}$ and by uniqueness of limit we have that $f=1_{\limsup_kE_{n_k}}$ a.e on $A$
Almost uniform convergence of a sequence $f_n$ to some $f$ implies that $f_n \to f$ pointwise a.e on $A$
So it is similar to the other modes of convergence.
Is this proof correct or am i missing something?
It seems easy enough to me. (easy off course if we have the implication about subsequences,because these implications are not so easy to prove)
Best Answer
If $1_{E_{n_k}}(\omega) \to f(\omega)$ then there is some $K$ and $v(\omega) \in \{0,1\}$ such that if $k \ge K$ then $1_{E_{n_k}}(\omega)=v$. Hence we can define $E$ ae. by $1_E(\omega) = v(\omega)$. It follows that $f=1_E$ ae.