# If $\mu$ is a complex measure, every set $E$ has $A \subset E$ so that $|\mu(A)| \ge \frac{1}{\pi}|\mu|(E).$

If $\mu$ is a complex measure on a $\sigma$-algebra $M$, show that every set $E \in M$ has a subset $A$ for which $$|\mu(A)| \ge \frac{1}{\pi}|\mu|(E).$$ The suggestion is as follows: Put $d\mu=e^{i\theta}d|\mu|$. Let $A_\alpha$ be the subset of $E$ where $cos(\theta - \alpha) > 0$. Show that $$Re[e^{-i\alpha}\mu(A_\alpha)]=\int_Ecos^+(\theta-\alpha) d|\mu|,$$ and integrate with respect to $\alpha$.

Show, by an example, that $1/\pi$ is the best constant in this inequality.

Thanks for the help.

From the definitions in the hint, we have:

$$e^{-i\alpha}\mu(A_\alpha) = \int_{E} \chi_{A_\alpha} e^{i(\theta - \alpha)}\,d|\mu|$$

Thus:

$$\left|\mu(A_\alpha)\right| \ge \Re\left(e^{-i\alpha}\mu(A_\alpha)\right) = \Re\left(\int_{E} \chi_{A_\alpha} e^{i(\theta - \alpha)}\,d|\mu|\right) = \int_E \cos^+(\theta-\alpha)\,d|\mu|$$

Choose $\alpha = \alpha_0$ to maximize the integral on the RHS. The value at $a_0$ is at least as large as the average:

$$\left|\mu(A_{\alpha_0})\right| \ge \frac{1}{2\pi} \int_{-\pi}^\pi \int_E \cos^+(\theta-\alpha)\,d|\mu|d\alpha$$

Use Fubini's theorem to swap integral signs on the RHS and obtain:

$$\left|\mu(A_{\alpha_0})\right| \ge \int_E \frac{1}{2\pi} \int_{-\pi}^\pi \cos^+(\theta-\alpha)\,d\alpha d|\mu|= \frac{1}{2\pi}\int_E 2 \,d|\mu| = \frac{1}{\pi}|\mu|(E)$$

To show that this is the best possible constant, consider the unit circle equipped with $d\mu = e^{i\theta} \, dm$ where $m$ is the Lebesgue measure. For any measurable set $E$ and $A \subset E$, pick $\alpha$ so that $e^{-i\alpha}\mu(A) \ge 0$ (always possible). I'll let you show that $\left|\mu(A)\right| \le \frac{1}{\pi}|\mu|(E)$.

• @CHG No problem. I also added an example to show that $1/\pi$ is the best possible constant. – Ayman Hourieh Mar 23 '13 at 17:09