Does the existence proof of $\mu$-completion require choice?

Let $(X,\mathfrak{M},\mu)$ be a measure space and $\mathfrak{M}^*=\{E\subset X|\exists A,B\in\mathfrak{M} \text{ such that} A\subset E \subset B \text{ and} \mu(B\setminus A)=0\}$.

How do i show that $\mathfrak{M}^*$ is closed under countable union?

Following is what i tried:

Let $\{E_n\}$ be a sequence of elements of $\mathfrak{M}^*$.

Define $I_n=\{(A,B)\in\mathfrak{M}\times\mathfrak{M}|A\subset E_n\subset B \text{ and} \mu(B\setminus A)=0\}$

Again, define $P_n=\{A\in \mathfrak{M}|(A,B)\in I_n\}$ and $Q_n=\{B\in \mathfrak{M}|(A,B)\in I_n\}$.

Then, $\bigcup P_n$ and $\bigcap Q_n$ are elements of $\mathfrak{M}^*$ for every $n\in\mathbb{N}$.

It can be shown that [If $C$ is measurable and $\bigcup P_n\subset C \subset E_n$, then $C\subset \bigcup P_n$], which means that there is no measurable set between $\bigcup P_n$ and $E_n$.

Is this $P_n$ and $Q_n$ are in $\mathfrak{M}$?

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Do you mean to ask whether $\bigcup P_n,\bigcap Q_n$ are in $\frak M$? – Asaf Karagila Jan 14 '13 at 8:04
@Asaf only if it is true, but it seems false now. My main interest is whether there exists a $\mu$-completion in ZF :) – Katlus Jan 14 '13 at 8:10
You mean a $\sigma$-additive completion. – Asaf Karagila Jan 14 '13 at 8:14
@Asaf Yes. I want to show that $\mathfrak{M}^*$ is a $\sigma$-algebra. So far, I have shown that $X\in\mathfrak{M}^*$ and [$E\in\mathfrak{M}^*\Rightarrow X\setminus E\in\mathfrak{M}^*$] – Katlus Jan 14 '13 at 8:19
It is trivially an algebra. Yes. – Asaf Karagila Jan 14 '13 at 8:25