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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)

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  • $\begingroup$ What does the $\limsup_k$ bit add? $\endgroup$ – copper.hat Oct 21 '19 at 15:01
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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.

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  • $\begingroup$ where am i wrong in my solution? $\endgroup$ – Marios Gretsas Oct 21 '19 at 15:18
  • $\begingroup$ @MariosGretsas: Nothing wrong, but the $\limsup_k$ bit suggests that you are misunderstanding something. You know that $\lim_k 1_{E_{n_k}}(\omega) = f(\omega)$ for ae. $\omega$, so I do not understand why you suddenly introduced the $\limsup_k$. Again, it is not incorrect, but seems entirely superfluous. The main point of the proof is that if $x_n \to x$ and $x_n \in \{0,1\}$ then we must have $x \in \{0,1\}$. $\endgroup$ – copper.hat Oct 21 '19 at 15:30
  • $\begingroup$ I just took limsup again to prove that $f$ is not just an indicator function of some set,but to find the set also which is $\limsup_kE_{n_k}$..but ok yes...i understand what you say. $\endgroup$ – Marios Gretsas Oct 21 '19 at 15:35
  • $\begingroup$ Yes, but the fact that $\lim_k$ exists is sufficient. $\endgroup$ – copper.hat Oct 21 '19 at 15:38

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