Upper and lower bound on $L^1$ norm purely in terms of measure

Question

Suppose $f$ is a measurable almost everywhere finite function on $\mathbb{R}^d$, and let$$E_n = \{x : 2^n \le |f(x)| < 2^{n + 1}\}, \quad n \in \mathbb{Z}.$$What is a non-trivial upper and lower bound on $\|f\|_{L^1}$ purely in terms of $m(E_n)$?

Answer 1

Let $E_\infty = \{x : f(x) = 0\}$. Then clearly $\int_{E_\infty} |f| = 0$. Since $f$ is a measurable almost everywhere finite function, we have$$\mathbb{R}^d = \left(\bigcup_{n \in \mathbb{Z}} E_n\right) \cup E_\infty \cup F,$$with $m(F) = 0$. By the monotonicity property of Lebesgue integration, we have$$\int_{\mathbb{R}^d} |f| = \sum_{n \in \mathbb{Z}} \int_{E_n} |f|+ \int_{E_\infty} |f| + \int_F |f| = \sum_{n \in \mathbb{Z}} \int_{E_n} |f|.$$Now for each $n \in \mathbb{Z}$, we have $2^n \le |f| < 2^{n + 1}$, so again by the monotonicity property of integration,$$2^n m(E_n) \le \int_{E_n} |f| \le 2^{n + 1} m(E_n).$$Sum over all $n \in \mathbb{Z}$, we get the following upper and lower bound on $\|f\|_{L^1} = \int_{\mathbb{R}^d} |f|$:$$\sum_{n \in \mathbb{Z}} 2^n m(E_n) \le \|f\|_{L^1} \le \sum_{n \in \mathbb{Z}} 2^{n + 1}m(E_n).$$