# Integrable functions as elements of closed absolutely convex hulls of precompact sets of indicator functions

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 $$L_1(\mu)$$?

• 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? May 1, 2020 at 15:35
• Dieter, yes, the closed absolutely convex hull, of course! Excuse me, I'll correct this. May 1, 2020 at 15:37
• 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? May 1, 2020 at 23:13
• @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 $C>0$. May 2, 2020 at 4:10

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 $$2=f(1)-f(-1)=\beta-\gamma\le |\beta|+|\gamma|\le 1$$.

• Why $\sum_{i=1}^{m_n}|\lambda_i^n|\le 1$? May 1, 2020 at 22:14
• 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 $L_1(\mu)$. May 2, 2020 at 7:01
• Are you saying that this is not true when $f$ has different signs at different points? May 2, 2020 at 7:04
• 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? May 2, 2020 at 18:59