1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Looks okay, except at the end. Given $\epsilon$ you define $\delta = \frac{\epsilon}{C}$. $\endgroup$ – Umberto P. May 12 '17 at 17:13
  • 2
    $\begingroup$ Really, a formal proof should start with your definition of $\delta$ based on $\varepsilon$ and $C$. But this scratch work is good. $\endgroup$ – Michael Burr May 12 '17 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.