Let $f$ be a complex-valued integrable function. Write the complex number $\int fd\mu$ in its polar form, letting $w$ be a complex number of absolute value 1 such that\begin{align}\int fd\mu = w\left|\int fd\mu\right|.\end{align}Then,\begin{align}\left|\int fd\mu\right| = w^{-1}\int fd\mu = \int(w^{-1}f)d\mu = \int\mathfrak{R}(w^{-1}f)d\mu \leq \int |f|d\mu.\end{align}
I have difficulty understanding why $\int(w^{-1}f)d\mu = \int\mathfrak{R}(w^{-1}f)d\mu$ is true. Could someone please help me out? Thanks a lot in advance!
I want to write $f=u+iv$ and $w=\cos\theta-i\sin\theta$. Then $w^{-1} = \cos\theta-i\sin\theta$ and\begin{align}w^{-1}f = (u\cos\theta+v\sin\theta)+i(v\cos\theta-u\sin\theta).\end{align}I couldn't see why $\mathfrak{I}(w^{-1}f) = v\cos\theta-u\sin\theta = 0$.
Let $(X,\mathscr{A},\mu)$ be a measure space. A complex-valued function $f$ on $X$ is integrable if its real and imaginary parts $\mathfrak{R}(f)$ and $\mathfrak{I}(f)$ are integrable; if $f$ is integrable, then its integral is defined by\begin{align*} \int fd\mu = \int\mathfrak{R}(f)d\mu + i\int\mathfrak{I}(f)d\mu.\end{align*}