Is it true that an open ball $U\subseteq L^p(X, \sum, \mu)$ centered at $0$ contains an indicator function of a positive measure?

Question

Let $(X, \sum, \mu)$ be a $\sigma$-finite measure space and $1\leq p<\infty$. It is known that the measurable simple functions are dense in $L^p((X, \sum, \mu)$.

In my research, I need to find an indicator function in an open ball $U\subseteq L^p((X, \sum, \mu)$ of radius $\epsilon$ centered at $0$.

Let $\delta<\epsilon$ and $A\subseteq X$ be a measurable set with $\mu(A)<\delta$. Is it true the indicator function $1_A$ is in $U$?

Answer 1

This is a a little too long for a comment. So I organize it as an answer.

First: In general case, without additional conditions, it may happen that the only indicator function in an open ball $U\subseteq L^p(X, \Sigma, \mu)$ of radius $\epsilon$ centered at $0$ is the indicator function of the empty set as we see in the following example:

Consider $(\Bbb N, 2^{\Bbb N}, \#)$ where $\#$ is the counting measure. Clearly $(\Bbb N, 2^{\Bbb N}, \#)$ is $\sigma$-finite. Take any $\varepsilon = \frac{1}{2}$. Then, for any set $S \subseteq \Bbb N$ and $p \in [1, +\infty)$, if $$ \left ( \int (\chi_S)^p d\#\right)^\frac{1}{p} = (\#S)^\frac{1}{p} < \frac{1}{2} = \varepsilon$$ then $\#S=0$. So $S=\emptyset$.

So the indicator function of the empty set is the only only indicator function in an open ball $U\subseteq L^p(X, \Sigma, \mu)$ of radius $\frac{1}{2} $ centered at $0$.

Second: On the other hand, given a measure space $(X, \sigma, \mu)$, if we assume the additional condition that

for every $\varepsilon >0$, there is a measurable set $A$ such that $0<\mu(A)<\varepsilon$ $\:\:\:(1)$

it is immediate that, taking a measurable set $A$ such that $0<\mu(A)<\varepsilon^p$, we have $$ \left ( \int (\chi_A)^p d\mu \right)^\frac{1}{p} = (\mu(A))^\frac{1}{p} < \varepsilon$$ and clealy $\chi_A \neq 0$ (because $\mu(A)> 0$).

Third: The condition $(1)$ is pretty simple. Another simple condition that implies condition $(1)$ is:

There are measurable sets with finite positive measure in $(X, \sigma, \mu)$ and there is no $\mu$-atom. $\:\:\:(2)$

It is easy to prove that $(2)$ implies $(1)$. But, of course, $(1)$ does not imply $(2)$.

We also have

There is $f \in L^1(X, \Sigma, \mu) \setminus L^\infty(X, \Sigma, \mu)$. $\:\:\:(3)$

Condition $(3)$ is actually equivalent to condition $(1)$.

Answer 2

Let $A$ such that $\mu(A)\le \epsilon^p$, $$\left\|\mathbf 1_A - \mathbf 0\right\|_p = \left(\int_{X} \left(\mathbf 1_{A}(x) - \mathbf 0(x)\right)\mathrm d\mu(x)\right)^{\frac1p} = \mu(A)^{\frac1p} \le \epsilon$$ so $\mathbf 1_{A} \in U$.