# a sequence of simple functions converges pointwise to a measurable function (Terence Tao)

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.

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 $$f_n(x) \leq f(x) . This shows that $$f_n(x) \to f(x)$$.
• Thanks. I guess that you mean $f_n(x) \le f(x) < f_n(x) + \frac1{2^n}$, right? – DEJABLUE Apr 7 at 13:00