Does $\frac{\lfloor 2^n f(x)\rfloor }{2^n}\vee(-n)$ is a simple function that converges to $f(x)$ when $f(x)<0$? Let denote $a\vee b:=\max\{a,b\}$ and $\lfloor y\rfloor$ the floor function of $y$. Let $f$ a measurable function. Let $x$ s.t. $f(x)<0$. Do we have that $$\varphi _n(x)=\frac{\lfloor 2^n f(x)\rfloor }{2^n}\vee (-n),$$
 is a simple function that converges to $f(x)$ ?

Indeed $\varphi _n(x)\to f(x)$, but someone told me that $\varphi _n$ is not a simple function, and I don't really understand why. Don't we have that $$\varphi _n(x)=\sum_{k=1}^{n2^n}\frac{-k}{n}\boldsymbol 1_{\{-\frac{k}{n}\leq f(x)<\frac{-k+1}{2^n}\}}(x)-n\boldsymbol 1_{\{f(x)<-n\}}(x).$$ 
Q1) Why isn't it a simple function ? I think it is.
Q2) If it's indeed a simple function, at the end, don't we have that $$\psi_n(x)=\left(\frac{\lfloor 2^n f(x)\rfloor }{2^n}\vee (-n)\right)\boldsymbol 1_{\{f(x)<0\}}+\left(\frac{\lfloor 2^n f(x)\rfloor }{2^n}\wedge n\right)\boldsymbol 1_{\{f(x)\geq 0\}},$$
is a sequence of simple function that converges to $f(x)$ ?
 A: According to your definition of a simple function, I assume that $f$ is a real-valued (but not necessrily measurable) function defined on a measurable space $(X,\Sigma)$. 
Q1. The answer depends on $f$. In general, a function $\varphi_n$ can be non-simple. For instance, take any measurable space $(X,\Sigma)$ with a non-measurable subset $A$ of $X$ (for instance, $X=\Bbb R$, $\Sigma$ is a family of all subsets $S$ of $X$ such that either $S$ or $X\setminus S$ is countable, and $A$ is a subset of $X$ such that both set $A$ and $X\setminus A$ are uncountable) and let $f(x)={\bf 1}_A(x)-2$, for each $x\in X$, where ${\bf 1}_A$ is the indicator function of the set $A$. Then $\varphi_2=f$ is not a simple function, because a set $f^{-1}(\{-1\})=A$ is not measurable. On the other hand, it is easy to see that for each $n\ge 1$ the function $\varphi_n$ is simple iff a preimage $f^{-1}\left(\left[\tfrac {k}{2^n},\tfrac{k+1}{2^n}\right)\right)$ is measurable for each integer $k$, such that $-n2^n<k<0$. The latter holds when $f$ is measurable.
Q2. As I understood, $f$ is a function from $X$ to $\Bbb R$, and $ a\wedge b:=\min\{a,b\}$. Then it is easy to check that for each $x\in X$ and each $n\ge |f(x)|$ we have $\psi_n(x)\le f(x)< \psi_n(x)+\tfrac 1{2^n}$, so a sequence $\{\psi_n(x)\}$ converges to $f(x)$.
The simplicity of the functions $\psi_n$ is similar to that in Q1. Namely, the range of $\psi_n$ is finite and for $n\ge 1$ it is measurable iff a preimage $f^{-1}\left(\left[\tfrac {k}{2^n},\tfrac{k+1}{2^n}\right)\right)$ is measurable for each integer $k$, such that $-n2^n <k<n2^n$ and both preimages $f^{-1}\left(\left(-\infty,-n+\tfrac{1}{2^n}\right)\right)$ and 
$f^{-1}\left(\left[n,+\infty\right)\right)$ are also measurable. These conditions hold when $f$ is measurable.
A: Answer to Q1
$\phi_n(x)$ is not subject to be a simple function. Define $$f(x)=\begin{cases}-1&,\quad x\in\Bbb Q\\-2&,\quad x\notin\Bbb Q\end{cases}$$since $f(x)$ is measurable (a similar problem can be found in Showing Dirichlet function is measurable) we have $$\varphi_n(x)=f(x)$$which is not simple but $\varphi_n(x)\to f(x)$. The most general constraint on $f(x)$ such that $\varphi_n(x)$ be a simple function is that the set of discontinuities of $f(x)$ must be countable. 
Answer to Q2
The answer is semi-affirmative. The function $\varphi_n(x)$ may or may not be a simple function just like what we said for Q1, but clearly tends to $f(x)$ since the first term of RHS tends to $f(x)$ when $f(x)<0$ and the second term is zero. Also the second term of RHS tends to $f(x)$ when $f(x)\ge 0$ and the first term is zero
