Problem on Integration: $\Bbb R-\Bbb C$ split and pull back of forms This post is not short. However I'm sure that a guy who good handle these concepts, could read and answer in five minutes. I only want to write my attempt, in order to understand where I'm wrong.
Let $\Omega\in\Bbb C$ be a domain; $\varphi:\Omega\to[-\infty,+\infty[$ upper semicontinous (i.e. $\varphi(z_0)\ge\limsup_{z\to z_0}\varphi(z)\;\;\forall z_0\in\Omega$).
I need to show, given $\bar\Delta_{z_0,r}\Subset\Omega$ (the unitary disk of radius $r$) that
$$
\varphi(z_0)\le\frac1{2\pi r}\int_{\partial\Delta_{z_0,r}}\varphi(s)\;ds
$$
implies
$$
\varphi(z_0)\le\frac i{2\pi r^2}\int_{\Delta_{z_0,r}}\varphi(z,\bar z)\;dz\wedge d\bar z\;\;.
$$
Now what everybody would do is to rewrite the first inequality as
$$
(2\pi t)\varphi(z_0)\le\int_{\partial\Delta_{z_0,t}}\varphi(s)\;ds
$$
and then integrate over $]0,r]$ wrt the variable $t$.
In this way LHS become easily $\pi r^2\varphi(z_0)$.
My problem is with RHS. Roughly I'd write
$$
\int_0^r\int_{\partial\Delta_{z_0,t}}\varphi(s)\,ds\,dt
=\int_{\Delta_{z_0,r}}\varphi(z)\,dz
$$
but I'm not sure it has some sense.
So I started form the other side, and this is what I got:
\begin{align*}
\int_{\Delta_{z_0,r}}\varphi(z,\bar z)\;dz\wedge d\bar z
&=-2i\int_{\Delta_{z_0,r}}\varphi(x,y)\,dx\wedge dy\\
&=-2i\int_{\Delta_{z_0,r}}\varphi(x,y)\,dx\wedge dy
\end{align*}
this could be seen in two ways: writing $dz\wedge d\bar z=(dx+idy)\wedge(dx-idy)=-2i(dx\wedge dy)$ or changing variable via the isomorphism $\beta:\Bbb R^2\stackrel{\simeq}{\to}\Bbb C$ defined by $(x,y)\mapsto(x+iy,x-iy)$.
What we have, then, is the integral of a $2$-form on a $2$-parametric manifold. So I consider now
$$
\alpha:]0,r[\times[0,2\pi[\longrightarrow\Delta_{z_0,r}\setminus\{z_0\}
$$
defined by
$$
(t,\theta)\longmapsto (\Re z_0+t\cos\theta,\Im z_0+t\sin\theta)
$$
from which we have
$$
-2i\int_{\Delta_{z_0,r}}\varphi(x,y)\,dx\wedge dy=\\
=-2i\int_{]0,r[\times[0,2\pi[}\varphi (\Re z_0+t\cos\theta,\Im z_0+t\sin\theta)
\underbrace{[\partial_t\alpha_1\partial_{\theta}\alpha_2-\partial_t\alpha_2\partial_{\theta}\alpha_1]}_{=\partial\alpha_1\wedge\partial\alpha_2(\partial_t\alpha,\partial_{\theta}\alpha)=t}\,dt\,d\theta\\
=-2i\int_0^rt\underbrace{\int_0^{2\pi}\varphi (\Re z_0+t\cos\theta,\Im z_0+t\sin\theta)\,d\theta}_{=:A(t)}\,dt
$$
And till here it seems (to me!) to have made no mistakes.
Now I thought I could work on $A(t)$. Using $\beta^{-1}$ I got
$$
A(t)=\frac i2\int_0^{2\pi}\varphi(z_0+te^{i\theta})\,d\theta
$$
The problem comes now: how to proceed?
I would change variable to this last integral $s=z_0+te^{i\theta}$: in this way it's ALMOST (and almost in Mathematics means wrong) equal to the wanted integral, except for a $ie^{i\operatorname{arg}(s)}$ or something similar (I erased it from my notebook).
Where am I wrong? How can I conclude?
Many thanks!
 A: You didn't quite edit correctly. The $i/2$ belongs in the inequality 
$$\varphi(z_0)\le \frac1{2\pi} \frac i2\int_{D(z_0,r)} \varphi(z)dz\wedge d\bar z.$$
Although I'm a huge fan of using $dz$ and $d\bar z$, I think it's easier just to use a polar coordinates double integral on this problem.
Given that $$\varphi(z_0)\le \frac1{2\pi r}\int_{\partial D(z_0,r)} \varphi (s)ds = \frac1{2\pi}\int_0^{2\pi}\varphi(z_0+re^{i\theta})d\theta$$
for all $0\le r\le R$, we proceed as follows:
\begin{align*}
\frac{R^2}2\varphi(z_0) &= \int_0^R \varphi(z_0) r\,dr \le \frac1{2\pi}\int_0^R\left(\int_0^{2\pi}\varphi(z_0+re^{i\theta})d\theta\right)r\,dr \\
&= \frac1{2\pi}\int_{D(z_0,R)} \varphi(z) r\,drd\theta = \frac1{2\pi}\frac i2\int_{D(z_0,R)} \varphi(z)\,dz\wedge d\bar z.
\end{align*}
This gives $\varphi(z_0) \le \displaystyle{\frac 1{\pi R^2}\frac i2\int_{D(z_0,R)} \varphi(z)\,dz\wedge d\bar z}$, as requested.
A: I can rewrite $A(t)$ as
$$
\int_0^{2\pi}\varphi(z_0+te^{i\theta})\,d\theta
$$
WITHOUT adding $-2i$ or $i/2$ factors (which are the $\Bbb R^2\longleftrightarrow\Bbb C$ passage) because I didn't change variable, I simply rewrote $(\Re z_0+t\cos\theta,\Im z_0+t\sin\theta)$ in another way.
In this way I came to
$$
-2i\int_0^rt\int_0^{2\pi}\varphi(z_0+te^{i\theta})\,d\theta\,dt\;\;.
$$
Now we change variable $s=z_0+te^{i\theta}$ from which $ds=|ite^{i\theta}|d\theta=td\theta$ (because $s$ is the element of arc). Thus the last integral is equal to
$$
-2i\int_0^r\int_{\partial\Delta_{z_0,t}}\varphi(s)\,ds\,dt\;.
$$
Hence
\begin{align*}
\frac{1}{-2i}\int_{\Delta_{z_0,r}}\varphi(z,\bar z)dz\wedge d\bar z
&=\int_0^r\int_{\partial\Delta_{z_0,t}}\varphi(s)\,ds\,dt\\
&\ge\int_0^r\varphi(z_0)2\pi t\,dt\\
&=\pi\varphi(z_0)r^2
\end{align*}
which concludes this little "odissey".
Thanks to all, once again I learned something new.
I'm going to bed, here it's 5am.
