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!


1 Answer 1


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 $\mu(E)=0$.

  • $\begingroup$ But why is $F$ measurable in the second case? That is what I wanted to ask. $\endgroup$ Dec 9, 2020 at 5:58
  • 1
    $\begingroup$ $E$ is the complement of $\bigcap_k \bigcup_n \bigcap_{m\geq n} \{x: |f_m(x)| <\frac 1k\}$. @AniruddhaDeshmukh $\endgroup$ Dec 9, 2020 at 6:01

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