# Modes of convergence and a.e equality with an indicator function.

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)

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

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.
• @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 $x \in \{0,1\}$. – copper.hat Oct 21 '19 at 15:30
• 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. – Marios Gretsas Oct 21 '19 at 15:35
• Yes, but the fact that $\lim_k$ exists is sufficient. – copper.hat Oct 21 '19 at 15:38