Completion and function Let $(X,\Sigma,\mu)$ a measure space and let $(X,\widetilde{\Sigma},\widetilde{\mu})$ its completion.
I want to prove that if $f:X\to\mathbb{R}$ is $\widetilde{\mu}$-measurable, then there exists $g:X\to\mathbb{R}$ $\mu$-measurable such that $f=g$ ($\widetilde{\mu}$-almost everywhere)
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How should I define $g$?
Each element of $\widetilde{\Sigma}$ is contained in a $A\cup B$, where $A,B\in\Sigma$ and $\mu(B)=0$. I want to define $g$ something like $g(x)=f(x)$ if $x\in A$ and $g(x)=0$ if $x\in B$. Of course I don't know how to do it in a formal way. 
Any suggestion? Thanks.

First, it suffices to assume $f : X \to [0,\infty)$ (why?).
I assume that you know that for every measurable function $f : X \to [0,\infty)$, there is a sequence $(f_n)_n$ of simple functions with $f_n \uparrow f$ (pointwise).
Here, every simple function $f_n$ is of the form $f_n = \sum_{i=1}^{m_n} \alpha_i^{(n)} \chi_{M_i^{(n)}}$ with measurable sets $M_i^{(n)}$. Here, measurability is to be understood as measurability w.r.t. the $\sigma$-algebra $\mathcal{A}$ with respect to which $f$ is measurable.
By what you have written, there is for each $M_i^{(n)} \in \tilde{\Sigma}$ a set $N_i^{(n)} \subset M_i^{(n)}$ that only differs up to a null-set from $M_i^{(n)}$.
Now, define $g_n := \sum_{i=1}^{m_n} \alpha_i^{(n)} \chi_{N_i^{(n)}}$ and set $g := \limsup_n g_n$.
Check that $g$ is $\Sigma$-measurable (easy) and that $f = g$ almost everywhere (use that countable unions of nullsets are nullsets).
