Question: I have a few questions regarding conjugates and moduli

1.) If $z=x+bi$, how would I simplify $\overline{z+a}$ or $\overline{z+i}$
Also, how would I simplify something like $\overline{2\overline{z}+5}$.

2.) Would $\overline{2} = 2$? In addition, would $\overline{3i}=-3i$?

3.) For the modulus, I have no idea how to simplify
$|3z-i|$, or $|z+2|$.

Thank you so much!

Answer: 
Assume that $X\_n, X$ are non-negative integrable RVs such that

$$ X\_n \to X \quad \text{a.s.}, \tag{1} $$
$$ \mathbb{E}[X\_n] \to \mathbb{E}[X] \tag{2}$$

hold. We also note that $\left|\max\{X, X\_n\} - X\right| \leq X\_n$. Then by the Fatou's Lemma,

\begin{align\*}
\mathbb{E}[X]
&= \mathbb{E}\left[\liminf\_{n\to\infty} (X\_n - \left|\max\{X, X\_n\} - X\right|)\right] \tag{by (1)} \\
&\leq \liminf\_{n\to\infty} \mathbb{E}\left[ X\_n - \left|\max\{X, X\_n\} - X\right| \right] \tag{by Fatou} \\
&= \mathbb{E}[X] - \limsup\_{n\to\infty} \mathbb{E}\left[ \left|\max\{X, X\_n\} - X\right| \right] \tag{by (2)} \\
\end{align\*}

This shows that $\limsup\_{n\to\infty} \mathbb{E}\left[ \left|\max\{X, X\_n\} - X\right| \right] = 0$, which in turn implies the desired $L^1$-convergence.
Sorry for the incorrect answer earlier. Let me provide a new solution here. The main result I apply here is:

> 
> Let $0<p<\infty$, if $(X\_n)$ is a sequence of random variables in $L^p$, and $X\_n \overset{\mathcal{P}}{\to}X$. Then the following are equivalent:  
> 
>  1. $|X\_n|^p$ is uniformly integrable;  
> 
>  2. $X\_n\to X$ in $L^p$;  
> 
>  3. $\mathbb{E}|X\_n|^p\to \mathbb{E}|X|^p<\infty$.
> 
> 
> 

Since $X\_n\to X$ a.s., then $\max\{X\_n, X\}\to X$ a.s.. From $\mathbb{E}(X\_n)\to \mathbb{E}(X)$ and $X\_n\ge 0$, by the result cited above, we have $X\_n\to X$ in $L^1$. Thus $\max\{X\_n, X\} = \frac{1}{2}(X\_n+X+|X\_n-X|)\in L^1$. To show $\max\{X\_n, X\}\to X$ in $L^1$. It suffices to show $\mathbb{E}\left(\max\{X\_n, X\}\right) \to \mathbb{E}(X)$, this is clear because
\begin{align}
\mathbb{E}(\max\{X\_m,X\}) = \frac{1}{2}\left[\mathbb{E}(X\_n)+\mathbb{E}(X)+\mathbb{E}|X\_n-X|\right] \to \mathbb{E}(X).
\end{align}
The result I cited here is from Theorem 4.5.4 in Kailai Chung's *A Course in Probability Theory*.