If $f:\mathbb R\to\mathbb R$ measurable and negative, there is a sequence $(\varphi _n)$ of simple function s.t. $\varphi _n\nearrow f$ Theorem: I know that if $f:\mathbb R\to\mathbb R$ is measurable and positive, there is a sequence of simple function $(\varphi _n)$ s.t. $\varphi _n(x)\nearrow f(x)$. I kniw that this result is not true if $f$ is not positive.


My question :

Let $f:\mathbb R\to\mathbb R$ measurable. Let $f\leq 0$ and $$\varphi _n(x)=\frac{\left\lfloor 2^n f(x)\right\rfloor }{2^n}\vee (-n),$$
where $\lfloor x\rfloor$ is the floor function and $a\vee b$ denote the maximum between $a$ and $b$. We have that $$\varphi _n(x)\to f(x)$$
and that $\varphi _n$ is increasing even since $$\varphi _n(x)=\frac{\lfloor 2^n f(x)\rfloor }{2^n}\leq \frac{2^{n}f(x)}{2^n}=\frac{2^{n+1}f(x)}{2^{n+1}},$$ 
and thus $\varphi _n(x)\leq \varphi _{n+1}(x) $.

Why is this theorem is not true if $f&lt;0$ ? The fact that there is a sequence of simple $\varphi _n(x)\nearrow f(x)$ looks true even if $f\leq 0$, no ?
It can't be true if $f$ is not bounded below. Because a simple function is certainly bounded below, and if $\phi_n$ increases to $f$ then $f\ge\phi_1$.
This result is true even for $f \leq 0$. What was probably meant is that you can't use this method to integrate functions that aren't positive almost everywhere.