If $\left|\frac{1}{\mu(E)}\int_{E} f d\mu\right| \leq M$ for all $E$, then $|f(x)|\le M$.

Question

Let $f$ be integrable and let there exist $M>0$, such that $$\left|\frac{1}{\mu(E)}\int_{E} f d\mu\right| \leq M$$ for every $E \in S$ with $0<\mu(E)<\infty$. Show that $|f(x)| \leq M $ almost everywhere relative to $\mu$.

I am still trying.

Answer 1

Suppose for a contradiction there is a set of positive measure $E$ on which $|f(x)| > M$ for all $x \in E$. Let $E_+ = \{x \in E: f(x) > M\}$ and $E_- = \{x \in E:f(x) < -M\}$, and note that $E = E_+ \cup E_-$ and $E_+ \cap E_- = \emptyset$. Thus at least one of $E_+$ and $E_-$ has positive measure; without loss of generality, assume $\mu(E_+) > 0$. Then:

$$\left|\frac{1}{\mu(E_+)} \int_{E_+} f d\mu \right| > \frac{1}{\mu(E_+)} \int_{\mu(E_+)} M d\mu = M,$$ which contradicts the assumption in (ii).