Let $\nu$ be a complex measure on $(\mathcal X,\cal M)$. Suppose that $L^1(\nu)=L^1(|\nu|)$ and $f \in L^1(\nu)$. Show that $$\left|\int f d\nu \right| \le \int \left|f\right| d\left|\nu\right|$$
My Attempt:
Step-1:
Suppose that $f=\chi_{E}$. Then by the definition of $|\nu|$, for $\mu=|\nu_r|+|\nu_i|, $we have $$|\nu(E)|=\left|\int_E g d\mu\right| \le \int_E\left|g\right|d\mu=|\nu|(E)$$
Hence this holds for simple functions. Now let $f$ be non-negative integrable function. Then there exists simple functions $0 \le \phi_1 \le \ldots\le\phi_n\le\ldots$ such that $\phi_n \to f$. Then
$$\left|\int \phi_n d\nu \right| \le \int \left|\phi_n\right| d\left|\nu\right|=\int \phi_n d\left|\nu\right|$$
Then $$\int \phi_n d\nu=\int \phi_n d\nu_r+i \int \phi_n d\nu_i=\left(\int \phi_n d\nu_r^{+}-\int \phi_n d\nu_r^{-}\right)+i \left(\int \phi_n d\nu_i^{+}-\int \phi_n d\nu_i^{-}\right)$$
Letting n $\to \infty$, we get that
$$\lim_{n \to \infty}\int \phi_n d\nu=\left(\lim_{n \to \infty}\int \phi_n d\nu_r^{+}-\lim_{n \to \infty}\int \phi_n d\nu_r^{-}\right)+i \left(\lim_{n \to \infty}\int \phi_n d\nu_i^{+}-\lim_{n \to \infty}\int \phi_n d\nu_i^{-}\right)=\left(\int f d\nu_r^{+}-\int f d\nu_r^{-}\right)+i \left(\int f d\nu_i^{+}-\int f d\nu_i^{-}\right)=\int f d\nu$$
Hence we have $$\left|\int f d\nu \right| \le \int \left|f\right| d\left|\nu\right|$$
Let $f$ be a real valued integrable function. Then $f=f^{+}-f^{-}$. So $$\left|\int f d\nu \right|=\left|\int f^{+} d\nu -\int f^{-}d\nu \right| \le\left|\int f^{+} d\nu\right| +\left|\int f^{-}d\nu \right| \le \int \left|f^{+}\right| d\left|\nu\right|+\int \left|f^{-}\right| d\left|\nu\right|=\int f^{+} d\left|\nu\right|+\int f^{-}d\left|\nu\right|=\int|f| d\left|\nu\right|$$
Now Let $f$ be a complex valued integrable function. Then there exists $\alpha \gt 0$ such that $$\int f d\nu =\alpha e^{i\theta}$$ Then $$\left|\int f d\nu \right|=\alpha=e^{-i\theta}\int f d\nu=Re\left(\int e^{-i\theta}f d\nu\right)=\int Re \left(e^{-i \theta}f \right) d\nu\le \int \left|e^{-i\theta}f\right| d|\nu|=\int |f| d|\nu|$$
Is this proof alright??
Thanks for the help!!