Help with a proof regarding simple functions. The question is 

If $f>g≥0$, then there exists non-negative measurable simple functions $f_k↗f$ s.t. $f_k≥g$ for all $k$.

My attempt. Using a theorem in my text book

For every non-negative function $f$, there exists a sequence $f_k$ of non-negative simple functions s.t. $f_k↗f$.

Suppose $g_k$ is a sequence of non-negative simple function s.t. $g_k↗g$. For any $k$, $f-g_k>0$, then there exists non-negative simple functions $h_j↗f-g_k$, thus $h_j+g_k↗f$ with respect to $j$.
But I got stuck here. Sine $g_k\le g$, I cannot conclude $h_j+g_k \ge g$. Hope someone can help. Thank you!
 A: I don't think this is possible. To see this, let
\begin{align*}
f(x)\equiv&\,\frac{1}{x},\\
g(x)\equiv&\,\frac{1}{x^2},\\
\end{align*}
for each $x\in[1,\infty)$. Clearly, $f>g>0$. I claim that there exists no simple measurable function $\varphi:[1,\infty)\to\mathbb R$ such that $f\geq\varphi\geq g$ pointwise (let alone a sequence of such functions). To see this, suppose, for the sake of contradiction, that $\varphi$ is a such a function. Then, there exist some $N\in\mathbb N$, some measurable subsets $A_1,\ldots, A_N$ of $[1,\infty)$, and distinct non-negative numbers $c_1,\ldots,c_N$ such that $$\varphi=\sum_{n=1}^Nc_n\mathbb 1_{A_n}.$$ It is without loss of generality to assume that $A_1,\ldots,A_N$ are not empty and disjoint. Because of this and the fact that $\varphi\geq g>0$, it must be the case that $c_1,\ldots,c_N$ are all positive and $A_1,\ldots,A_N$ form a partition of $[1,\infty)$. Let $\varepsilon\equiv\min\{c_1,\ldots,c_N\}>0$. Because of $\lim_{x\to\infty}f(x)=0$, there exists some $x\in[1,\infty)$ such that $f(x)<\varepsilon\leq\varphi(x)$, which contradicts $f\geq\varphi$.
