Show $|\int f d \nu| \leq \int |f| d |\nu|$

Question

Let $\nu$ be a complex measure and $f \in L^1( \nu)$. Prove that $$\left|\int f d \nu\right| \leq \int |f| d |\nu|$$

Here $|\nu|$ is the total variation of $\nu$.

I managed to prove that the result holds when $f$ is real-valued by the usual argument. However, when $f$ is complex-valued I have trouble. I tried

$$\left|\int f d \nu\right|^2= \left|\int \Re (f) d \nu + i \int \Im (f) d \nu\right| ^2$$ $$= \left|\int \Re(f)d \nu\right|^2 + \left|\int \Im(f) d \nu\right|^ 2$$ $$\leq \left(\int |\Re(f)|d |\nu|\right)^ 2 + \left(\int |\Im(f)|d |\nu|\right)^2$$

and I want to end up with $$=\left(\int (\Re(f)^2 + \Im(f)^2)^{1/2} d |\nu|\right)^2$$ $$\leq \left(\int |f|d |\nu|\right)^2$$

How can I complete the estimation? I feel like I'm missing easy.

Answer 1

Using the polar decomposition of a complex measure one can argue similarly as in the case of complex Lebesgue integrals: There is a measurable, real-valued function $\theta$ such that $$ \int f d\nu = \int f e^{i\theta} d|\nu| \, . $$ It follows that $$ \left| \int f d\nu \right | = \left| \int f e^{i\theta} d|\nu| \right | = e^{-i\alpha} \int f e^{i\theta} d|\nu| = \int f e^{i\theta - i \alpha} d|\nu| $$ for some real number $\alpha$. The right-hand side is a non-negative real number, so this is equal to $$ \operatorname{Re} \left( \int f e^{i\theta - i \alpha} d|\nu| \right) = \int \operatorname{Re} (f e^{i\theta - i \alpha}) d|\nu| \le \int |f e^{i\theta - i \alpha}| d|\nu| = \int |f| d|\nu| \, . $$

Answer 2

If $f$ is a simple function, say, $f=\sum\limits_{k=1}^{n} c_k\chi_{E_k}$ then the inequality is clear. I general there exist simple functions $f_j$ such that $|f_j| \leq |f|$ and $f_j \to f$ point-wise. Apply the inequality for each $f_j$ and take limit as $j \to \infty$. The fact that $\int f_j d\nu \to \int f d\nu$ can be seen by writing $\nu$ as a linear combination of four positive finite measures and applying DCT. Also $\int |f_j| d\|\nu| \to \int |f|d |\nu|$ by DCT.