Properties of plurisubharmonic functions In book: (Oxford science publications._ London Mathematical Society monographs, new ser., no. 6) Maciej Klimek -Pluripotential theory -OUP (1992)
Theorem 2.9.14 (ii):

Let $\Omega$ be an open subset of $\mathbb{C}^n$. If $\Omega$ is connected
  and $\{u_j\}_{j \ge 1} \subset PSH(\Omega)$ is a decreasing sequence,
  then $u=\lim_{j \to \infty}u_j \in PSH(\Omega)$ or u $\equiv - \infty$.

Here's my solution:
$\bullet$ We'll prove that: $u$ is upper - semicontinuous in $\Omega$. 
Because $u_j \in PSH(\Omega), \forall n \ge 1$ 
$\implies u_j$ is upper - semicontinuous in $\Omega$. 
Then $\forall \alpha \in \mathbb{R}: \{z \in \Omega: u_j(z)<\alpha \}$ be an open set in $\Omega$.
We have: $\left \{ z \in \Omega: u(z)<\alpha \right \}=\bigcup_{n\ge 1}\left \{ z \in \Omega: u_j(z)<\alpha \right \}$ be an open set in $\Omega$.
$\bullet$ We'll prove that: $\forall a \in \Omega,b\in \mathbb{C}^n$ such that $\left \{ a+\lambda b: \lambda \in \mathbb{C},\left |\lambda  \right | \le 1 \right \}\subset \Omega$ then 
$$u(a)\le \dfrac{1}{2\pi}\int_{0}^{2\pi}u(a+e^{i\theta}b)\mathrm{d}\theta$$
Since $u_j \in PSH(\Omega), \forall j \ge 1 $ then $\forall a \in \Omega,b\in \mathbb{C}^n$:
$$u_j(a)\le \dfrac{1}{2\pi}\int_{0}^{2\pi}u_j(a+e^{i\theta}b)\mathrm{d}\theta$$
Because $\{u_j\}_{j \ge 1} \subset PSH(\Omega)$ is a decreasing sequence, $\lim_{j\ge 1}u_j=u$ then 
$$\lim_{j \ge 1}u_j(a)\le \lim_{j \ge 1} \left ( \dfrac{1}{2\pi}\int_{0}^{2\pi}u_j(a+e^{i\theta}b)\mathrm{d}\theta \right )$$.
Therefore
$$u(a)\le \dfrac{1}{2\pi}\int_{0}^{2\pi}u(a+e^{i\theta}b)\mathrm{d}\theta .\,\,\ \square $$
Theorem 2.9.14 (iii):

Let $\Omega$ be an open subset of $\mathbb{C}^n$.  If $u : \Omega \to
 \mathbb{R}$, and if $\{ u_j \} \subset PSH(\Omega)$ converges to $u$
  uniformly on compact subsets of $\Omega$ , then $u \in PSH(\Omega)$.

I want to prove Theorem 2.9.14 (iii), i have thought about it. I mean
$\bullet$ We'll prove that: $u$ is upper - semicontinuous in $\Omega$.  
$\bullet$ We'll prove that: $\forall a \in \Omega,b\in \mathbb{C}^n$ such that $\left \{ a+\lambda b: \lambda \in \mathbb{C},\left |\lambda  \right | \le 1 \right \}\subset \Omega$ then 
$$u(a)\le \dfrac{1}{2\pi}\int_{0}^{2\pi}u(a+e^{i\theta}b)\mathrm{d}\theta$$.
However, I have trouble when I'm trying to use $\{ u_j \} \subset PSH(\Omega)$ converges to $u$ uniformly on compact subsets of $\Omega$ (by assumptions), I've forgotten the definition, maybe . 
EDITED
Now, I remember it :)
A sequence $u_j$ converges uniformly on $K \subset \Omega$ mean $\forall \epsilon >0, \exists N \in \mathbb{N}$ such that $\forall z \in K$ and $\forall j \ge N$ we have $\sup_{z \in K} \left| f_j(z)-f(z)\right | < \epsilon$
Can anyone help me! Thanks.
 A: Your proof of 2.9.4 (ii) looks ok. You should perhaps mention that you're using the monotone convergence theorem at the end.
For the corresponding statement about uniform convergence, you can use a very similar argument to establish the sub-mean value inequality as the one you use for 2.9.4 (ii). Just replace the monotone convergence theorem with uniform convergence.
Remember that being plurisubharmonic is a local property, so it's enough to show the inequality for analtic discs $a+\lambda b$ whose image is relatively compact in $\Omega$. To be pedantic, that's what you're assuming in (ii) as well in order to be sure that you can integrate over the boundary of the analytic disc.
What remains is to show that the limit $u$ is upper semicontinuous. Fix a point $z_0 \in \Omega$ and a compact set $K \ni z_0$. Let $\varepsilon > 0$ and take $n$ so large that $\|u_n-u\| < \varepsilon$ on $K$ (this is possible by uniform convergence). Since $u_n$ is usc, we can find a neighborhood $V$ (which we can take inside $K$ by shrinking $V$ if necessary) of $z_0$ such that $u_n(z) \le u_n(z_0) + \varepsilon$ for  $z \in V$.
By uniform convergence,
$$
u(z) \le u_n(z) + \varepsilon \le u_n(z_0) + 2\varepsilon \le u(z_0) + 3\varepsilon
$$
for all $z \in V$. This shows that $u$ is upper semicontinuous at $z_0$, but $z_0 \in \Omega$ was arbitrary and we are done.
