I am not a specialist in measure theory, so excuse me if this is simple.

Let $\mu$ be a finite measure on a set $X$ (for example, the Lebesgue measure on $[0,1]$). Integrable functions on $X$ can be defined as limits (in different senses, in particular, in some definitions as uniform limits, but this does not play an important role in what I want to ask) of summable step functions $$ f=\lim_{n\to\infty}\sum_{i=1}^\infty\lambda_i^n\cdot\chi_{A_i^n}, $$ where $\lambda_i^n\in{\mathbb C}$, $A_i^n\subseteq X$.

Is it possible to control the set $\{\lambda_i^n\cdot\chi_{A_i^n};\ i,n\in{\mathbb N}\}$ in this equation so that it would not be "too big"?

For example,

Suppose $f\in L_1(\mu)$ and $|f(x)|\le 1$, $x\in X$. Do there exist a sequence of measurable sets $A_i\subseteq X$ such that

1) the set of indicators $\{\chi_{A_i};\ i\in{\mathbb N}\}$ is totally bounded (or, what is the same, precompact) in $L_1(\mu)$, and

2) $f$ belongs to the closed absolutely convex hull of the set $\{\chi_{A_i};\ i\in{\mathbb N}\}$ in $L_1(\mu)$?

  • 1
    $\begingroup$ Without going into detail: I think the problem is assumption 2). Is it allowed that $f$ belongs to the "closed" absolute convex hull? What about the usual elementary three-step definition of the integral: First define $\int f d\mu$ for simple functions (as your sums, but with finite sum), then for non-negative functions and then for arbitrary quasiintegrable functions? $\endgroup$ May 1, 2020 at 15:35
  • $\begingroup$ Dieter, yes, the closed absolutely convex hull, of course! Excuse me, I'll correct this. $\endgroup$ May 1, 2020 at 15:37
  • $\begingroup$ You ask first that the set $\{\lambda_i^n\cdot\chi_{A_i^n} ; i, n \in \mathbb N\}$ be not too big, but your proposed 'bigness' conditions seem to ignore the coefficients $\lambda$. Is that intentional? $\endgroup$
    – LSpice
    May 1, 2020 at 23:13
  • $\begingroup$ @LSpice I had in mind that if the condition $|f(x)|\le 1$ is not fulfilled, then $\chi_{A_i}$ must be multiplied by some $C>0$. $\endgroup$ May 2, 2020 at 4:10

1 Answer 1


Attempt number 2. Consider the case $f\ge 0$.

For $\alpha\in[0,1]$ let $A_{\alpha}=\{x\in X, f(x)\ge \alpha\}$, which is measurable. For $n\in \mathbb{N}$ define $f_n=\frac{1}{n}\sum_{k=1}^{n}\chi_{A_{\frac{k}{n}}}$. It is easy to see that $f_n\le f\le f_n+ \frac{1}{n}$, from where convex combinations of $\chi_{A_{\frac{k}{n}}}$ converge to $f$. We only have to show that $\mathcal{A}=\{\chi_{A_\alpha}\}$ is totally bounded.

Let $\alpha, \beta\in [0,1]$. If $\alpha\ge\beta$ then $A_{\alpha}\subset A_{\beta}$, from where $\|\chi_{A_\alpha}-\chi_{A_\beta}\|_1=\mu(A_{\beta})-\mu(A_{\alpha})$. Similarly, if $\alpha\le\beta$, then $\|\chi_{A_\alpha}-\chi_{A_\beta}\|_1=\mu(A_{\alpha})-\mu(A_{\beta})$, and so in both cases $\|\chi_{A_\alpha}-\chi_{A_\beta}\|_1=|\mu(A_{\alpha})-\mu(A_{\beta})|$.

Hence, $\chi_{A_\alpha}\to\mu(A_{\alpha})$ isometrically maps $\mathcal{A}$ into a totally bounded space $[0,\mu(X)]$, and so $\mathcal{A}$ is itself totally bounded.

In the case $f$ is complex-valued, decompose $f=g-h+i(p-q)$, where $0\le g,h,p,q\le 1$, and so each of them belongs to the closed convex hulls of a totally bounded collection of indicators.

However, you cannot get general $f$ into a single convex hull. Consider $X=\{-1,1\}$ and $f$ - identity map. There are just three non-zero indicators available: $\chi_{X}, \chi_{\{1\}}, \chi_{\{-1\}}$. Assume that $f$ is in the (closed) absolute convex hull of these three elements. If $|\alpha|+|\beta|+|\gamma|\le 1$, and $f=\alpha\chi_{X}+\beta\chi_{\{1\}}+\gamma\chi_{\{-1\}}$, then $1=f(1)=\alpha+\beta$, and $-1=f(-1)=\alpha+\gamma$, and so $2=f(1)-f(-1)=\beta-\gamma\le |\beta|+|\gamma|\le 1$.

  • $\begingroup$ Why $\sum_{i=1}^{m_n}|\lambda_i^n|\le 1$? $\endgroup$ May 1, 2020 at 22:14
  • $\begingroup$ erz, I did not understand your reasoning about the mapping $\chi_{A_\alpha}\mapsto\mu(A_\alpha)$, but independently on this, the family $\{\chi_{A_\alpha}\}$ is totally bounded in $L_1(\mu)$ because in each sequence $\{\chi_{A_{\alpha_i}}\}$ we can choose a subsequence $\{\chi_{A_{\alpha_{i_k}}}\}$ that has limit in $L_1(\mu)$. $\endgroup$ May 2, 2020 at 7:01
  • $\begingroup$ Are you saying that this is not true when $f$ has different signs at different points? $\endgroup$ May 2, 2020 at 7:04
  • $\begingroup$ @SergeiAkbarov I've edited the answer to address both of your comments $\endgroup$
    – erz
    May 2, 2020 at 18:12
  • $\begingroup$ erz, isn't it necessary to prove that the image of the map $\chi_{A_\alpha}\mapsto \mu(A_\alpha)$ is closed, for applying what you say about the isometry? $\endgroup$ May 2, 2020 at 18:59

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