Let $(X,A,\mu)$ be a measure space and $\int_{X}f\; d\mu = \int_{X}|f|\; d\mu$ with $f \in \mathscr{L}^{1}(\mu)$.
Then $f$ can be written as $f = u + iv$, where $u,v$ are real-valued functions, and it follows from the condition above that $\int_{X}v\;d\mu = 0$. At this point, of course, it can't be inferred that $v$ is a zero function ($v=0 \;\mu\text{-a.e.})$
Also from the condition above we know that $\int_{X}u\; d\mu = \int_{X}\sqrt{u^2 + v^2}\;d\mu$.
Allegedly from here on it can be shown that $v$ is a zero function by defining sets $E_n := \{x\in X \mid v^2(x)>\frac{1}{n}\}$ and finding an estimate for the last integral to show that $\mu(E_n)=0$ for every $n$.
I can't seem to find a decent definition for $E_n$ nor a good estimate for the integral. Do any of you know how this can be done?
Maybe it is simpler with $$E_n:=\bigl\{\, x\in X\mid v^2(x)>(|u(x)|+\tfrac1n)^2-u(x)^2\,\bigr\}.$$ Then on $E_n$, $$\sqrt{u(x)^2+v^2(x)}>|u(x)|+\frac1n$$ so that $$0=\int_X\sqrt{u^2+v^2}\,\mathrm d\mu-\int_X u\,\mathrm d\mu\ge\int_{E_n}\left(\sqrt{u^2+v^2}-|u|\right)\,\mathrm d\mu\ge \frac1n\mu(E_n) $$ and so $\mu(E_n)=0$. As $\bigcup_n E_n=\{\,x\in X\mid v(x)\ne 0\,\}$, the claim follows.