Approximating bounded measurable functions by simple functions from above

It is a standard result in measure theory that measurable functions can be approximated arbitrarily close from below by simple functions. More precisely,

a measurable function $$f$$ is the limit of a sequence of simple functions $$f_n$$ with $$|f_n| \leq |f|$$ for all $$n$$.

The proof is usually done by constructing the following sequence of functions: $$f_n(x) = \begin{cases}\frac{k-1}{2^n} & \text{if } \frac{k-1}{2^n} \leq f(x) < \frac{k}{2^n} \text{ for some } k = 1, 2, \dots, n2^n \\ n & \text{if } f(x)\geq n\end{cases}$$

I would like instead to approximate measurable functions from above. It is clear that this cannot be done for arbitrary measurable functions, since if the function $$f$$ is unbounded, we cannot even find a single simple function $$s$$ such that $$s$$ is above $$f$$. Hence, I only consider bounded, measurable functions. My hope is to prove the following.

A bounded, measurable function $$f$$ is the limit of a sequence of simple functions $$f_n$$ with $$|f| \leq |f_n|$$ for all $$n$$.

My idea for a proof is to adapt the previous construction to this case. Assume that $$f$$ is a measurable function that is bounded by $$M$$. Then construct $$f_n(x) = \begin{cases} 0 & \text{if } f(x) = 0 \\ \frac{k}{2^n} & \text{if } \frac{k-1}{2^n} < f(x) \leq \frac{k}{2^n} \text{ for some } k = 1, 2, \dots, n2^n \\ M & \text{otherwise}\end{cases}$$

I feel that this should do the trick, but my analysis is very rusty, and I am not sure how to go about proving (or disproving) it.

Can someone help me show that is a correct approach, or show that the desired theorem does not hold?

• It seems that in the construction you are only looking at nonnegative $f$, right? Then it looks okay to me. Also you can take simple functions $g_n$ converging to $M-f$ with $g_n\leq M-f$, and $f_n:=M-g_n$ will converge to $f$ with $f\leq f_n$. – drhab Jun 23 '17 at 8:53

Take simple functions $g_n$ converging to $M - f$ with $g_n \leq M - f$ and define $f_n := M - g_n$. Then $f_n$ converges to $f$ and $g_n \leq M - f$ implies $g_n - M \leq - f$, so $M - g_n \geq f$, which is the same as saying $f_n \geq f$, so the $f_n$ satisfy the required conditions.