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 $L\_1(\mu)$?
> 
> 
> 



Integrable functions as elements of closed absolutely convex hulls of precompact sets of indicator functions
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 $2=f(1)-f(-1)=\beta-\gamma\le |\beta|+|\gamma|\le 1$. 


