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Lemma: Let $\Omega$ be a measurable subset of $\mathbb{R}^n$, and let $f : \Omega \to \mathbb{R}$ be a measurable function. Suppose that $f$ is always non-negative, i.e., $f(x) \ge 0$ for all $x \in \Omega$. Then there exists a sequence $f_1, f_2, f_3, ... $ of simple functions, $f_n :\Omega \to \mathbb{R}$, such that the $f_n$ are non-negative and increasing, $$0 \le f_1(x) \le f_2(x) \le f_3(x) \le ... \text{for all $x \in \Omega$}$$

and converge pointwise to $f$:

$$\lim_{n \to \infty} f_n(x) = f(x) \text{ for all $x \in \Omega$}.$$

Prove this Lemma. (Hint: set $$f_n(x) : = \sup\{\frac{j}{2^n} : j \in \mathbb{Z}, \frac{j}{2^n} \le \min (f(x), 2^n)\},$$ i.e., $f_n(x)$ is the greatest integer multiple of $2^{-n}$ which does not exceed either $f(x)$ or $2^n$. You may wish to draw a picture to see how $f_1, f_2, f_3$, etc. works. Then prove that $f_n$ obeys all the required properties.

The same question is answered here, but I think that it uses somewhat different approach (or I am failing to observe the connection between these two). I am trying to solve this questions, sticking to the hint given, but I am struggling with it. I appreciate if you give some help.

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Hints: $f_n(x)$ takes only the values $0,\frac 1 {2^{n}},\frac 2 {2^{n}},... \frac {2^{2n}} {2^{n}}$ and it takes these values on measurable sets. It is clear that $f_n(x)$ is increasing. Verify that $f_n(x) \leq f(x) <f_n(x)+\frac 1{2^{n}}$. This shows that $f_n(x) \to f(x)$.

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  • $\begingroup$ Thanks. I guess that you mean $f_n(x) \le f(x) < f_n(x) + \frac1{2^n}$, right? $\endgroup$ – DEJABLUE Apr 7 at 13:00
  • $\begingroup$ Yes, corrected. $\endgroup$ – Kavi Rama Murthy Apr 7 at 13:02

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