# Pointwise and Pointwise almost everywhere convergence of monic simple functions

I am doing a course on measure theory in which we have defined monic simple functions on a measure space $$\left( X, \mathcal{B}, \mu \right)$$ as follows:

Definition: A monic simple function is of the form $$f = \alpha \cdot \chi_E$$, where E is a measurable set.

Now, we consider a sequence of monic simple functions $$f_n = \alpha_n \cdot \chi_{E_n}$$ such that for all $$n \in \mathbb{N}$$, we have $$\alpha_n > 0$$ and $$\mu \left( E_n \right) > 0$$. I want to look at its pointwise and pointwise almost everywhere convergence to the zero function. We have claimed that

1. $$f_n \to 0$$ pointwise if and only if $$\alpha_n \to 0$$ or $$\limsup\limits_{n \to \infty} E_n = \emptyset$$.
2. $$f_n \to 0$$ pointwise a.e. if and only if $$\alpha_n \to 0$$ or $$\mu \left( \limsup\limits_{n \to \infty} E_n \right) = 0$$.

Here,we have defined $$\limsup\limits_{n \to \infty} E_n = \bigcap\limits_{n \in \mathbb{N}} \bigcup\limits_{m \geq n} E_m$$.

My ideas so far:

To prove (1), if we consider $$f_n \to 0$$ pointwise, we may either have $$\limsup\limits_{n \to \infty} E_n = \emptyset$$ or it is non-empty. In case it is empty, there is nothing to prove. Therefore, we consider the case when, $$\limsup\limits_{n \to \infty} E_n \neq \emptyset$$. Then, there is some $$x \in \limsup\limits_{n \to \infty} E_n$$. This would however, mean that for each $$n \in \mathbb{N}$$, there is some $$m \geq n_0$$ such that $$x \in E_m$$. Also, pointwise convergence gives for each $$\epsilon > 0$$ some $$n_0 \left( x \right) \in \mathbb{N}$$ such that $$\left| f_n \left( x \right) \right| < \epsilon$$.

These two information together tell us that there is some $$m \geq n_0 \left( x \right)$$ such that $$\left| \alpha_m \right| < \epsilon$$. How should we conclude that for all $$m \geq n_0 \left( x \right)$$, we have $$\left| \alpha_m \right| < \epsilon$$?

The converse is pretty simple, and I am through it.

Now, to prove (2), I have assumed that (1) is indeed true. So, if we assume that $$f_n \to 0$$ pointwise a.e., then again, we have either $$\mu \left( \limsup\limits_{n \to \infty} E_n \right) = 0$$ or $$\mu \left( \limsup\limits_{n \to \infty} E_n \right) \neq 0$$. In the first case, there is nothing to prove. In the second case, we conclude that $$\limsup\limits_{n \to \infty} E_n \neq \emptyset$$ (since it is a non-null set), and therefore $$f_n \to 0$$ pointwise (everywhere) on this set. Then, from the above result, because the set is non-empty, we conclude that $$\alpha_n \to 0$$. Is this reasoning correct? I feel that there is some problem towards the end of the reasoning, although I cannot see it clearly.

Also, in the converse, if one assumes that $$\mu \left( \limsup\limits_{n \to \infty} E_n \right) = 0$$, how should we claim that $$f_n \to 0$$ pointwise a.e? If we start with a set $$E := \left\lbrace x \in X | f_n \left( x \right) \not\to 0 \right\rbrace$$, we should prove that $$\mu \left( E \right) = 0$$. Till now, I am able to prove that $$E \subseteq \limsup\limits_{n \to \infty} E_n$$. However, does this imply (using the monotonocity of the measure) that $$\mu \left( E \right) = 0$$? If not, then how do we prove that $$E = \limsup\limits_{n \to \infty} E_n$$?

Thank you for your help and suggestions!

The first part is actually false. Let $$E$$ be a set of positive measure, $$E_n=\emptyset$$ for $$n$$ odd and $$E_n=E$$ for $$n$$ even. Let $$\alpha_n=n$$ for $$n$$ odd and $$\alpha_n=\frac 1 n$$ for $$n$$ even. Then $$\alpha_n$$ does not tend to $$0$$ and $$\lim \sup E_n$$ is not empty even though $$f_n(x) \to 0$$ for every $$x$$.
In the second part, yes, $$E \subseteq F$$ and $$\mu (F)=0$$ does imply $$\mu(E)=0$$.
• But why is $F$ measurable in the second case? That is what I wanted to ask. Dec 9, 2020 at 5:58
• $E$ is the complement of $\bigcap_k \bigcup_n \bigcap_{m\geq n} \{x: |f_m(x)| <\frac 1k\}$. @AniruddhaDeshmukh Dec 9, 2020 at 6:01