# Prove $\forall \epsilon >0 \exists \delta > 0$ s.t. if $E \subseteq \Omega$ is a set with $\mu(E) < \delta$ then $\int_E |h| d\mu < \epsilon$.

for this question, I would just like to know whether or not my proof makes sense.

Suppose $h \in \mathcal{L}_{\infty}(\Omega)$. Prove that for any $\epsilon >0$, $\exists$ $\delta > 0$ s.t. if $E \subseteq \Omega$ is a measurable set with $\mu(E) < \delta$ then $\int_E|h| d\mu < \epsilon$.

$||f||_{\infty}=\inf \{C \geq 0: |f(x)| \leq \text{for } \mu-\text{almost all } x \in X \}$.

$\int_E|h|d\mu \leq C \int_E 1 d\mu = C \int_E 1 d\mu(E) = C\mu(E)$

If $\mu(E) < \delta$ then we can define $\epsilon = C \delta$ to give the desired inequality.

• Looks okay, except at the end. Given $\epsilon$ you define $\delta = \frac{\epsilon}{C}$. – Umberto P. May 12 '17 at 17:13
• Really, a formal proof should start with your definition of $\delta$ based on $\varepsilon$ and $C$. But this scratch work is good. – Michael Burr May 12 '17 at 17:14