# Question about formula for total variation of complex measure from Real Analysis of Folland

Let $\nu$ be a complex measure on $(X, \mathcal{M})$. If $E \in \mathcal{M}$, define:

• $\mu_1(E) = \sup\{\sum_1^n{|v(E_j)|}:n \in N, E_1, ..., E_n$ disjoint$, E = \bigcup_1^n{E_j}\}$
• $\mu_2(E) = \sup\{\sum_1^{\infty}{|v(E_j)|}:E_1, E_2,...$ disjoint$, E = \bigcup_1^{\infty}{E_j}\}$
• $\mu_3(E) = \sup\{|\int_E{fdv}|: |f| \le 1\}$.

Prove that $\mu_1 = \mu_2 = \mu_3 = |\nu|$

This problem is exercise 21, chapter 3 from Real Analysis of Folland. I followed the structure in his book, and up to now, I can prove that $\mu_1 \le \mu_2 \le \mu_3$ and $\mu_3 = |\nu|$. So the only point I stuck right now is to prove that $\mu_3 \le \mu_1$. I think about approximating $f$ using simple functions. But $f$ is a complex function, so the approximation is quite complicated. Anyone has any suggestion? Thanks so much for your help, I really appreciate.

• Since $|\int_E fd\nu| \leq \int_E |f|d\nu$, you are really taking the sup over non-negative real-valued functions (in the definition of $\mu_3$) – Prahlad Vaidyanathan Aug 27 '15 at 2:37
• Hm, your inequality is only right for positive measure, not completely right for complex measure. I think for complex measure, it should be $|\int_E{dv}| \le \int_E{|f|d|v|}$ – le duc quang Aug 27 '15 at 2:39
• Yes, you are right, of course. Some work is needed. – Prahlad Vaidyanathan Aug 27 '15 at 2:42
• We can write $f = f_r^+ - f_r^- + i(f_i^+ - f_i^-)$. But the operator $|.|$ makes it harder to approximate each one function with a simple function. Do you have any idea, @PrahladVaidyanathan – le duc quang Aug 27 '15 at 2:45
• Does it even make sense to consider complex $\nu$ in this case? Surely $\mu_1,\mu_2,\mu_3$ are positive measures? – Jason Aug 27 '15 at 2:49

Approximation by simple functions is the way to go. Fix $\epsilon>0$ and let $f$ be such that $|f|\le1$ and $\left|\int_Ef\, d\nu\right|\ge\mu_3(E)-\frac\epsilon2$. Now choose a simple function $\varphi=\sum_{j=1}^na_j\mathbf1_{E_j}$ (with $E_j$ disjoint and $\bigcup E_j=E$) such that $\left|\int_E f\, d\nu-\int_E \varphi\, d\nu\right|\le\frac\epsilon2$. It follows that $$\mu_3(E)\le\left|\int_Ef\, d\nu\right|+\frac\epsilon2\le\left|\int_E\varphi\, d\nu\right|+\epsilon=\left|\sum_{j=1}^na_j\nu(E_j)\right|+\epsilon\le\sum_{j=1}^n|\nu(E_j)|+\epsilon\le\mu_1(E)+\epsilon,$$ where the second last inequality follows since necessarily $|a_j|\le1$. Since $\epsilon>0$ was arbitrary we are done.
• I don't think you can choose a function $\phi$ like that when $f$ is complex function. As far as I know, the approximation like that can only applied for a measurable "positive" function. With complex function, I think you still need to split out into 4 functions like what I did. Did I miss anything? – le duc quang Aug 27 '15 at 3:49
• @leducquang It is enough to decompose $f$ into the real and imaginary parts and then, for each part, decompose it in the positive and negative parts. So you will have four positive functions $Re(f)^+$, $Re(f)^-$, $Im(f)^+$ and $Im(f)^-$. Each of them can be aproximated by a simples function: $\phi_{R,+}$, $\phi_{R,-}$ $\phi_{I,+}$ and $\phi_{I,-}$. It is easy to show that $f$ is approximated by $\phi=(\phi_{R,+} - \phi_{R,-}) + i(\phi_{I,+} - \phi_{I,-})$. – Ramiro Aug 27 '15 at 16:08
• @leducquang Yes, in my comment above $\phi$ is a complex-valued simple function. The general definition of simple functions does not require them to have values in $\mathbb{R}$. When working with compex-valued measurable functions, it is natural to consider complex-valued simple functions. – Ramiro Aug 27 '15 at 17:33